Okay, there's one exercise that was a little more difficult than the rest. Maybe I'll go through that. I don't think I assigned this one.

Did I assign that? Yes? Okay, let's do that one.

Okay, that the limit of a n is a, says this. Given epsilon bigger than zero, there is a capital N, so that a n is between a minus epsilon and a plus epsilon for all little n bigger than or equal to capital N. Do you agree with that statement?

That's what it means to have limit a. Alright? So because of that, this is the limit we want to treat. There's an a n in there, so let's replace a n on one side by a plus epsilon, the other side by a plus epsilon.

Do you agree with that? Okay, and what does this side do as n goes to infinity?

This converges to something, this one, e to the a minus epsilon. And this converges to e to the a plus epsilon.

Okay?

So, maybe you remember, do you know the terms lim inf and lim sup? No? Okay. Well, if this limit exists, what does it have to be between?

Right. It has to be between this and that, right? So the limit, let's assume the limit exists.

And this is true for every epsilon bigger than zero. No matter what epsilon I pick, this holds in the limit.

And what can I do with epsilon now? I can let epsilon go to zero, because the two n's squeeze down to e to the a.

Right? And any question on this? Okay, technically speaking, I should say, for those who do know this language, this holds.

And then you let epsilon go to zero, and then you conclude the lim inf is equal to the lim sup, and that means the limit exists and is their common value, which must be e to the a.

So for more rigorous treatment, if you've had these terms, I think you might have had them in calculus, when you're talking about limits of sequences. But anyway, I think the idea is clear from here. And then this was supposed to be used in another exercise, on the extreme order statistic for uniformly distributed random variables.

And I forget which number that was. 29? Okay, number 29.

So we take u1 through un to be iid uniformly distributed on the interval from zero to one.

And then, after you take your sample, you call the largest one that you see,

u sub parenthesis n. Okay, that's the max of the, that's the largest observation. And the exercise was to find the CDF, first to find the CDF for this. So let's find the CDF for that.

That was the easy part of the exercise. Does everybody know by now what this stands for? Okay, cumulative distribution function.

Well, the CDF for any random variable is this, if you put the random variable here, it's the probability that the random variable is less than or equal to x.

It's a function of x. It's this function of x we're interested in. And this says what? Un less than or equal to x. Un is the largest of all the observations. So if the largest is less than or equal to x, well, the largest is less than or equal to x if and only if all of them are less than or equal to x.

And these are independent, so this becomes, since these commas mean n or intersection,

this becomes a product of the probabilities of u1 less than or equal to x, of u2 less than or equal to x, et cetera. And I could put probably u2 less than or equal to x here, and then a bunch of dots and probably un is less than or equal to x there. But what does this I mean in IID?

Identically distributed. They have the same distribution function. So that you'd repeat the same thing over and over again, but how many times do you repeat it? N times. So we get that to the nth power. And what is this?

Well, depends on x. If x is bigger than 1, it's 1. Why is that?

What's the CDF for a uniformly distributed random variable on an interval from 0 to 1 look like? It's that, right? 0 until you get to 0, and it's the function x, f of x equal x, and then it's 1.

Why is it 1? Because if it's a uniformly distributed random variable on the interval from 0 to 1, the random variable has to have the value less than or equal to 1. So the probability that it's less than or equal to a number bigger than 1 is 1. If x is less than 0, then this probability is 0.

All the numbers are between 0 and 1, so they can't be less than 0. The maximum can't be less than 0. Otherwise, if x is between, then this thing is x, and we raise it to nth power.

Okay, let's find the expected value of un.

I'll put 0 in there because I know the density will vanish off the interval from 0 to 1.

Why? What's the density? Here's the CDF. The density is the derivative of this. What's the derivative of the constant function? 0. What's the derivative of this constant function? 0. So the PDF is 0, but for values of x less than 0 and for values of x bigger than 1.

You should have here the integral from minus infinity to infinity x times the density of un, but the density vanishes off the unit interval, so I don't have to go from 0 to 1. And what's the density between 0 and 1? It's the derivative of this. That would be n times x to the n minus 1.

And so inside here we get x to the n. The integral of x to the n is x to the n plus 1 over n plus 1, and we evaluate between 0 and 1.

The value at 1 is 1 over n plus 1 times n, and the value at 0 is 0. So there's the expected value of the max. Now let's find the variance of the max.

Well, that would be the expected value of the maximum squared minus the square of the expected value. We just computed this. It's here.

How do we compute this? Well, here we're computing the expected value if we put x there. What will we put if we're doing squared? We put x squared there.

OK, and this becomes integral from 0 to 1 n times x to what power? From here we get x squared. From here we get x to the n minus 1.

So that would be an x to the n plus 1 altogether? OK, so here's the CDF, and then the PDF is the derivative of that.

So we get x to the n plus 1 when we multiply these two together.

And then minus n divided by n plus 1 squared. And this becomes n over n plus 2 minus n over n plus 1 squared.

And let's combine this. We have to find common denominator. Common denominator would be n plus 2, n plus 1 squared.

We have to multiply this by this denominator. And we have to multiply n squared by this denominator.

And what does this become?

When I multiply the first term out here, I get n squared plus 2n plus 1. I multiply that by n, so I get n cubed plus 2n squared plus n.

And here I get minus n cubed minus 2n squared. For a grand total of what? Just n is left.

So there's the variance of un.

OK, so far so good? Now, the exercise says something about the distribution of the standardized max.

What's the standard normal distribution? Have we used that term? Has the book used that term? What do you think the standard normal would be? Mean 0, variance 1. So a standardized random variable means you somehow transform it so it has mean 0 and variance 1.

How can you transform a random variable so it has mean 0? Subtract the mean. So we'd subtract n over n plus 1. Now, how can you make a random variable have variance 1? Yeah, exactly. Divide by square root of the variance.

OK, so we'll take un, u parenthesis n, subtract that and divide by the square root of this.

Try to compute this. Less than or equal to x.

For simplicity, let me call this sigma n squared.

Well, if this number is between 0 and 1, we know what this is. It's what? It's that number to the power n. So if, and if it's bigger than 0, say.

Now let's put back in this for sigma n squared.

This should not be sigma n squared here, sorry. It should be square root of that, right? It's the square root of sigma n squared. So I better put square roots here. I do that because I can erase the 2 on the board, but I don't think you can erase the 2 in your notes.

Is that right now? OK, I'll start again. OK, so I should have square root of n over n plus 2,

and then x over n plus 1, plus n over n plus 1.

Can you see that? So the way to get from here to here is add a 1 here in the numerator and subtract a 1.

If I add a 1, n plus 1 over n plus 1 is 1. If I subtract a 1, that gives me minus 1 over n plus 1, which I put over here. OK? Now what was the exercise we did just a moment ago?

It was if a n goes to a, then 1 plus a n over n to the n goes to e to the a.

We almost have that here. This would be the a n. What does that converge to? Does that numerator converge to anything? What does this ratio n over n plus 2 do as n goes to infinity?

It goes to 1. So the numerator here goes to x minus 1. This numerator goes to x minus 1. And the denominator, is that n?

Not quite. It's a quibble that it's not, but we can fix that. Let's replace this with n. But we want to be honest. We want to keep the same quantity we had before. So I'll put an n here, and then in the numerator I'll put n over n plus 1.

So I'm going to rewrite this as, first I'm going to put the 1 because over there I put 1 first. Over n to the n. And then I'll put n over n plus 1 here times square root n over n plus 2 x minus 1.

And that's equal to this. See this n plus 1? That's in the denominator here, right? So that means it's the same as being down here. And when I have an n up here and an n down there, what do they do?

They cancel. So that gives that thing. Now, this is the a n. And what does that do as n goes to infinity? n over n plus 1 goes to 1.

n over n plus 2 goes to 1. So this whole numerator goes to x minus 1. So what does this expression go to then? What does this probability converge to? This converges to e to the x minus 1. So long as this is true, this should hold in the limit.

Let's see what this says. Mu n is n over n plus 1, so this should say that x is less than or equal to 1 minus mu n over sigma n.

And what does this do as n goes to infinity? 1 minus mu n is, what's 1 minus mu n? That's converging to 1, right? So 1 minus mu n is converging to 0. In fact, 1 minus mu n would be n plus 1 over n plus 1 minus n over n plus 1 is 1 over n plus 1, right?

This n plus 1 and this one down here will cancel.

It's n plus 1 squared, leaving n over n plus 2 square root. And what does this do as n goes to infinity? It goes to 1. So this is good as long as x is less than or equal to 1.

And the CDF for x bigger than that would be 1. So limiting CDF for the standardized max, what does this do when x is equal to 1?

e to the 0 or 1. So this CDF looks like what?

When x goes to minus infinity, this goes to 0. It dies off exponentially. It goes up to 1 at 1 and then it's flat. Turns out, if you ever go into insurance, this kind of random variable comes up a lot, the maximum.

I think I mentioned earlier why would an insurance company be interested in the maximum of a sequence of random variables. Well, the claims they receive on their policies are random variables. And the maximum would be the largest payout they have to make.

And so they want to charge enough money so that they can cover the largest payout they have to make. So they want to know the distribution of the largest payoff. n is the number of claims they've received. So they probably are thinking about a large number of claims over a period of years.

So they want to know how this behaves. So what they do is they have some distribution for the claims. It may not be uniform. It could be something else. And they subtract the mean and they divide by the square root of the variance. And it turns out that this thing always has a limit and there are three kinds of limits it can have.

There are three different limiting CDFs you can get depending on what distribution you start with. These are called the extreme value distributions. This is just one example of one of the extreme value distributions. You could start with another underlying distribution instead of uniform.

You would get a different one. But in the limit, there are only three possible limits that you can get. So they're very interested in the distribution of the largest claim that they can get. And it turns out that in the limit, there are only a couple different kinds of limits you can get. This is a very famous theorem in something called extreme value statistics.

And this is all in the area of risk management. People are risk averse. Some people are risk averse. How many of you think you're risk averse? How many of you go on the roller coaster ride at the Grey Ghost at Knott's Berry Farm?

Anybody go on that? So you're not risk averse. Or perhaps you invest your money in unwise companies. Or perhaps you go to Las Vegas and you're going to win.

But then there are some people who are very risk averse and they want to get rid of their risk. They sell it to somebody else and there's a whole industry. And this whole industry of risk management, or helping people who are risk averse, uses a lot of probability theory.

Okay, so that's it for those exercises. This is probably harder than anybody. Probably nobody was able to get this, but that's okay. So there will be a new set of exercises posted today.

They'll be due Friday. It'll be on chapters 6 and 7. And I think that'll be the end of the course. I don't remember what the syllabus said we're going to do. But I think we'll finish 6 and 7. Okay, so last time we discussed chi-squared distributions.

If z is normal with mean 0 and variance 1, then I'll define it to be u.

u equals z-squared is called a chi-squared with one degree of freedom.

If z1, z2, zn are independent, normal 0, 1 random variables,

then u sub n, I'll call it, equals z1-squared plus z2-squared plus zn-squared is chi-squared with n degrees of freedom.

These come up in a lot of statistical analyses.

Some other common distributions arising from the normal are, if m and n are non-negative integers,

and u is chi-squared with m degrees of freedom, v is chi-squared with n degrees of freedom, and u and v are independent, that's called the fmn distribution.

Ratios are useful statistics to analyze data.

Now last time we did mention that the chi-squared is actually a gamma.

We believe it's gamma of n over 2 and 1 half. That means this is the alpha, that's a lambda.

I haven't done a lot of computations where we compute PDFs or CDFs for ratios,

so let's do that now with this one.

So that would be the CDF differentiated, so we want to differentiate this thing.

So that means we should try to figure out what this is. Rather than write the derivative every time, I'm just going to compute this first and then I'll differentiate afterwards.

Let's multiply through by v over n. Does that change this sign at all? What were u and v? Chi-squares with m and n degrees of freedom respectively. What were chi-squares?

Sums of squares of normals. So chi-squared random variables are always non-negative or positive. So when I multiply through by v over n, I don't change the direction here at all.

And now I'm going to multiply through by little m. I'm going to write this m over n v x.

Using the independence of u and v, I could get to this.

Maybe I should take a couple steps here.

This is saying that the pair u, v is in a certain set. The pair has to lie below the line in u, v space that has slope m over n x.

That's this line, right? u less than or equal to that means... Well, maybe I have the wrong one. u less than or equal to that means I guess we're above the line.

The first variable has to be smaller than this multiple and the second. So the probability that u is less than or equal to m over n x, v would be the probability that the pair u, v is in that set A.

Maybe I'll index it by m over n x. So this is the set A, m over n x. Is that too quick? At this set, well, if the pair u, v is up here,

is it true that u is less than or equal to m over n v x?

Here's the fixed value of v. If I go along this line here, horizontal line with constant y coordinate v, when I start out here, is u bigger than m over n x, v or smaller? When u is 0, it's smaller and it's smaller all the way up until I get to that line there.

So this is the right region. And to compute a probability like this, you integrate the joint density over that region.

But they're independent.

So the joint density is a product of the individual densities. And now what we should do is set limits of regression to describe where we're integrating.

So our v can go from 0 to infinity.

And for a fixed value of v, u goes from here to m over n x, v. So this is the integral, v going from 0 to infinity, u going from 0 to m over n x, v,

f sub capital U of little u, f sub capital V of little v, du is the integral, dv is the outer one. Now let me kick this guy out.

And what is this integral? It's the integral of the density for capital U from 0 to something. What's that called? Three letters.

Last letter is f. CDF. That's the CDF for U evaluated at the upper limit. So I'm going to write that first. This is the CDF for U, so that's the probability that U is less than or equal to that value.

Okay? So that's what I wrote here. So always remember this.

If you want to compute the probability of something about a pair of random variables, you need their joint density or joint distribution or joint probability mass function. And you express whatever this restriction is as a set and you integrate the CDF over that set

or you sum the probability mass function over that set. So here that set was this. So this should be automatic every time. Alright, so we got to this point. And we wanted to differentiate.

So let's differentiate that expression. Well, here's the only place X appears, so we differentiate this.

This is the CDF for U, so when we differentiate, we should get the PDF. But we have to use the chain rule.

We're differentiating with respect to X, so we'll get this thing outside. So this will become integral from 0 to infinity. This PDF at M over NVX, because that's the derivative of this function of that variable. But then we have to multiply by the derivative of this with respect to X, which would be M over NV,

and then we have the PDF for V. Okay? So that's the chain rule. Right? This is capital F sub U at M over NVX.

So if we want to differentiate that with respect to X, we get capital F U prime at M over NVX times the derivative of what appeared inside with respect to X, that's M over NV. Okay? And this is the PDF now. This is little f.

Okay? And then we have to integrate this from 0 to infinity. But you won't have to do any integration. Let's substitute what the PDFs are now. I wrote them on the board.

They're going to be constants out in front. First one, I guess, will be M over N from here. And then we take the, evaluate the PDF for U at M over NVX.

Now U has these constants out in front, so let's put those out in front.

And then V will have one half to the N over gamma of a half. And then we take the value of this at, we raise that to the M over 2 minus 1.

And then we raise V to the N over 2 minus 1.

I promise you, it's not as bad as it looks. Okay. Now what in here doesn't get, doesn't have a V?

Well, this doesn't have a V. Let's pull that out, and the X doesn't have a V. Let's pull that out. Okay?

Now out in front, we have an M over N. Well, let's see. Let's take care of the halves first. We have a half to the, whoops, I forgot, N over 2 here. We have a half to the M plus N over 2. We have a gamma of one half squared.

Here's an M over N. But we also have an M over N to the M over 2 minus 1. Yeah?

Oh, you're right. Let's see. No? Oh, it's here. You're right. I didn't put it, I left out the V. Thanks. I left out the V. Yeah. I forgot to put this V down here.

Okay. So, M over N to the M over 2 minus 1. And here's an M over N to the first power, so we'll get altogether M over N to the M over 2.

And what's the, so I think that's all of the constants. Oh, well, I forgot this one. X to the M over 2 minus 1.

N times integral, zero to infinity. V to some power. What power?

Here's the first power. Here's to the M over 2 minus 1. Here's to the N over 2 minus 1. I add all that up. 1 plus M over 2 minus 1 plus M over 2 minus 1. That is V to the M plus N over 2 minus 1.

And then E to the minus V.

Okay. So, it's late. Let's take a break and come back in 10 minutes and finish it.

Okay. So, let's continue. But amend your notes. I've put gamma of a half here. It should be gamma of M over 2. Remember the gamma density is lambda to the alpha over gamma of alpha. X to the alpha minus 1.

E to the minus lambda X. So, that should be gamma of M over 2 here. Gamma of N over 2 there. So, that shows up here. It should have gamma of M over 2. Gamma of N over 2. And when I was evaluating the density for U at this,

I did put that in M over NVX to the M over 2 minus 1 for the power. But I forgot to do that in the exponential part. So, this should not be. This should be M over NVX here.

I forgot to put the, I put V in for the exponential part. So, I should be evaluating this at M over NVX.

So, that should go in here. It should also go in the exponent. I forgot to put it in the exponent. This is the density for U. Not to evaluate it here. I did it correctly there, but I forgot to put it in the exponent. So, that'll show up when I do the exponents.

I have an E to the minus V over 2 and then this. So, if I add the exponents, I'll add minus V over 2 M over NVX and minus V over 2.

I can factor out the minus V over 2. I get 1 plus M over NVX times V over 2. Or, maybe I'll write it this way, minus 1 plus M over NVX over 2 times V.

Okay? That's what happens when I combine these two exponentials. So, once again, I forgot to put M over NVX in the exponent part of the density for U. I put it in the power part, but you also have to put it in the exponent part.

And correct the densities for U and V to have a gamma of M over 2, not a gamma of a half. It's M over 2 here. And for V, N over 2. Gamma of N over 2. Okay? And that brings us to this here. And the last part I have to carry over from there is E to this power.

DV. Okay? So, how do we do this integral? We use the fact that this is a density, right?

And in that, what is the alpha? Alpha is M plus N over 2. And lambda is one half 1 plus M over NVX.

You compare. Alpha is M plus N over 2. Lambda is half times quantity 1 plus M over NVX.

And so, this becomes, let me put this down here.

and here I'd get lambda to the alpha in the denominator that'd be 1 half 1 plus m over n x raised to the power alpha right which is m plus n over 2

this is the lambda to the alpha another way of saying this here is that integral from 0 to infinity x to the alpha minus 1 e to the minus lambda x dx is gamma of alpha

over lambda to the alpha so I've just put the lambda to the alpha here with lambda equal half times the quantity 1 plus m over n x alpha equal m plus n over 2 and up here should go gamma of alpha well you can

combine the x terms but that's the density so there are a couple of steps, key steps one is if you're computing the probability

distribution function for a ratio you write that if the denominator is always a non-negative random variable you can multiply through by that and not change the direction of inequality and then you have to express this as the probability that the pair uv lies in some region

in this case would be above a line and then you integrate the joint density for the pair over the region above the line and if they're independent the joint density factors and the rest was just collecting terms busy work and then using the fact that you know

that the gamma density is a density and then you don't have to do any integration it's kind of ugly I have to admit it doesn't look very nice okay so another

distribution that comes up is called the t sometimes referred to as the student t distribution this is called the student t distribution

sometimes a student doesn't refer to people like yourselves it refers to

pen name of a statistician who worked for Guinness and he had to work with small samples and to analyze data from small samples he came up with this

statistic here we'll study this later I'll give it to you as a challenge to compute the density for this that would follow the same steps we've just done how do you compute the density you can write down the distribution function

what would that look like you'd have a what instead of this ratio you'd have a z here and a root u over n here you could multiply through by the root u over n so you'd have z less than or equal x root u over n this is the pair then that means the pair

z u must lie in a certain region in space write down the integral over that region of the joint density what's the joint density going to be for z and u it's going to be a product of the individual densities and then hope for some miracle

and then there will be one but i won't do it the density is written down in the book but see if you can derive it yourselves

okay so let's consider the following suppose x1

through xn are independent random variables well for now i don't need independent the sample mean is just this

they're average and the sample variance i always forget yeah right i want to make sure it's one over n minus one

here n minus one goes here the reason for

that is if the if the x's are

independent or let's say iid with variance common variance equal to sigma squared then the expected value of s squared is sigma squared this is supposed to be

an estimate for sigma squared if i have a one over n here i wouldn't get sigma squared there

i'd get n minus one over n times sigma squared if s squared is supposed to estimate some parameter namely sigma squared if the expected value of the estimator is equal to the true value then we call that an unbiased statistic so we normalize by one over n minus one here

to get an unbiased statistic up here um what would the expected value of x bar equal if we're in this situation again if these are identically distributed then the expected value of x bar is mu

why is that well what's the expected value of a sum of random variables

it's the sum of the expected values so what do we get here in the sum we get n times mu but then divide by n you get mu okay so this is supposed to be an estimator for mu and its expected value when you take the expected expectation of the randomness you get mu so that's called an unbiased estimator

its expected value is equal to the parameter it's supposed to be estimating so let's examine the case where we have normal random variables suppose these are

normal mean zero variance one random variables which is often the case because of the central limit theorem many samples and applications can be assumed to be

normal random variables also we have the fact that Poisson distribution when properly centered and scaled is close to a normal random variable and binomial random variables when we subtract the mean divide by the square to the variance that's going to be close to

normal so a lot of the samples you get are assumed to be normal random variables in that case x bar is independent of the random vector x1

minus x bar x2 minus x bar x and minus x bar it's kind of strange x bar appears in both places

yet they're independent so one way to prove independence and independence is look at the joint density in short factors another way is to look at the moment generating function

and show that it factors into a product if it does remember then use the uniqueness property of moment generating functions a moment generating function can only come from one distribution so if it comes from a distribution where the x bar part factors away from this part then you

know they're independent so let's look at the moment generating function for this for the vector value case we have n plus one random variables here so we need n plus one real variables and here's what we do we look at e to the t naught x bar plus t1 x1 minus x bar

plus t2 x2 minus x bar plus etc t n x n minus x bar and what we're going to do is we'll show that this factors into and that will be enough to show

independence okay okay let's recall this fact and

let's start with the left hand side okay and what's x bar what do we get

here what's the coefficient of um there's

there's there are a bunch of x ones in here right what's the coefficient of x1 in this expression in the exponent what do we get for the coefficient of x1 from here well x1 is the first term in the sum it's multiplied by

one over n and then we multiply also by t naught over n right so maybe i'll just work with the exponent for a moment or two here i got t naught over n

um here i won't get a what do i get here for the coefficient of x1 is that right and what do i get all along here the next term would be

t2 x2 minus x bar so i'll get a minus t2 over n times x1 from here then minus t3 over n etc and then here i get a minus tn over n all times x1

and what do i get for the coefficient of x2 from here i get a t naught over n from here i get a minus t1 over n from here i'll get a t2 times 1 minus 1 over n

then i get a minus t3 over n etc all the way up to minus t sub n over n times x2 and i keep going like that until i get down to the nth one okay

on the diagonal i get t1 times 1 minus 1 over n then t2 times 1 minus

1 over n and all the way down to here right let's recombine

there are terms with just like t1 t2 etc right so i get t sum i equal 1 to n ti times xi

well let's see what do i get times everyone has a t naught over n right

every one of them has a t0 over n every one of them has a that for x1 i get a t1 for x2 i get a t2 for x3 i get a t3 for x and i get a tn so i get a plus ti and then i get a minus

t sub i over n minus uh what some of you some j equal 1 to n tj over n i'm going to write that this way 1

over n sum j equal 1 to n tj

so let's do this one first if i have t naught over n times xi and i add up i going from 1 to n what do i get i get it i can follow up the t naught right

and then i get a 1 over n times the sum of the xi's that would be x plus the sum i equal 1 to n ti xi and then minus

what um this thing i could call t bar so we could rewrite it this way this is

not x bar is it but it's n times x bar so i combine that with this but uh actually that's not a that's not as

useful as i i would like i think i'll just stick with this let's just stop here okay we'll use this one okay so you don't need these what i wrote is right but we'll stop here okay now let's decide what kind of random variable is this we have a sum of independent normal

random variables what kind of random variable is t naught over n plus ti minus t bar xi that's a normal random variable what's its mean zero and its variance is

t naught over n plus ti minus t bar squared

if i have a normal random variable x is normal if x is normal means zero and variance one what kind of random variable is a times x what's the expected value of this zero still what's the expected value what's the variance the variance would just be the expected value of this squared

that would be a squared okay so we just get this squared so what kind of random variable would this whole sum be here they're independent normals the sum of independent normals is normal this is a normal random variable what's the expected value of

this each sum and as i take the value zero so the whole sum would have expected value zero and what about the variance of this

the variance of the sum is the sum of the variances for independent random variables so the variance would be the sum of these things now let's examine what's what this

variance is if you have a a plus b squared it's a squared plus

2 ab plus b squared so i just did that now what's the first term here all of these are the same so i just multiply this by n

that would be t naught squared over n what about this when i add up this what do i get when i add up the ti's i get

n times t bar when i add up these what do i get well they're all the same i get n times this term

but i'm subtracting this from this so what does this what does this become zero this part is zero okay so this random variable up there is

normal with this variance and mean zero okay so and that's what goes in this exponent this random variable is normal with mean zero and variance that so this is equal to

e to the mu is zero sigma squared is that thing over there and so

so this is so i have one here one times this random variable this is this is my this is now my x i'm taking t equal one so i get just the half the variance t naught squared over n or 2n and then times e to the

one half sum i equal one to n ti minus t bar squared so i find that the expected value of this thing the moment

generating function for this complicated vector form is this what is the expected value of e to the t naught x bar can i get that from this

is t naught x bar equal to what i see there for some choice of t1 through tn

yeah what choice take all the t i's to be zero except t zero then this will be zero that'll be zero that'll be zero what do you get here then zero so this would be

okay and what is this part that's if i take if i take t naught to be zero i that's what i get here and what if i take t naught to be zero i

just get this so what's the moral of the story the moment generating function for this is the product of this moment generating function in that one so what does that mean it means this random variable x bar and the vector x1 minus x bar through

x n minus x bar are independent okay and the miracle was right here

the moment generating function for normal looks like this it has this particular form so when we're

computing the variance what happened the t naught part separated from the ti minus t bar part and we're taking exponential of this so that means it factors that's that's why the factoring took place because this term cancelled

okay so x bar is independent of this when you have independent normal samples all right so this implies a very important result x bar and s squared are independent

s squared remember was one over n minus one

sum i equal one to n x i minus x bar squared this is a function of you know use g of x1 minus x bar x2 minus x bar x n minus x bar right

it's a it depends on what these differences s squared is a function of these differences x bar is independent of this vector this is a function of the vector that's independent of x bar so a function of

if you have two random variables y and x are independent that implies y and g of x are independent okay so we're just taking a function of these things and they're independent of x bar

so so when you're dealing with sample means and sample variances they are independent random variables if your sample is normal it's not true if it's not normal okay so we can stop there today have a good weekend