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(20 points) Suppose that sequence of measurable functions fn Xx[0,&] is given such tat < fn+(z) < fn(r) for all _ eX Assume further that fi € L'...

Question

(20 points) Suppose that sequence of measurable functions fn Xx[0,&] is given such tat < fn+(z) < fn(r) for all _ eX Assume further that fi € L'(X,e) i.e Jx f dp < &. Show thatlim +00J, dp Ks duHint: Use the convergence theorems! Bonus question (10 points): Show by counterex- ample that the conelusion is false if f 4 L'(Xp).

(20 points) Suppose that sequence of measurable functions fn Xx[0,&] is given such tat < fn+(z) < fn(r) for all _ eX Assume further that fi € L'(X,e) i.e Jx f dp < &. Show that lim +00 J, dp Ks du Hint: Use the convergence theorems! Bonus question (10 points): Show by counterex- ample that the conelusion is false if f 4 L'(Xp).



Answers

Conditioning on Full Information. Let (?, P) be a finite probability space and let F be the power set. Show that
any random variable X is F -measurable. Conclude that E[X | F ] = X.

Yeah. So in this problem we need to sure that any bounded decreasing sequence of real numbers converges. So we are given about a decreasing sequence. We named that A. N. And we need to show that A. N. Can gorgeous to what limit? We don't care about this. Okay, so let's start with the proof, send the sequence in is bounded in particular. It does bounded below that is the set A. And such that and comes from natural numbers is a bounded subset five. Have we known that are has completeness property also known as and you'll be property that is least upper bound property. And this is equal into G. L. B. Property also known as greatest lower bound property, basically this means, but if any subset of R is bounded below, then it must have an infamy. Um So in fema of the scent and coming from and coming from national numbers is some real number. Let's call this L. So our claim is like the sequence N. Can we just to L now, since L is the greatest No, it bound after said of elements of sequences of the terms of sequences. So for any absolutely. Didn't see role as minus epsilon, which is strictly less than N. I'm sorry, L plus epsilon, which is strictly created. An L cannot be by lower bound. So there exists some. And in particular there exists some term A N. That is less than L plus of silence. But since in is a decreasing sequence we have a N less than equal to capital and for every and greater than or equal to capital. And so in particular we have and less than L plus a silent for every angry to an end no sense. And is can feel them. We have less than equal to a N. For every n natural numbers. In particular, l minus epsilon, which is strictly less than L, would also be less than a N for every angrier than capital. And so we have this inequality right here and this inequality right here and we combined them to find that for every epsilon created in zero, there exists a national number and such side for every end. After that comes after that natural number added minus upside and is less than a. In which is less than L. Plus of silent. That means A. In the sequence A in Canada. Just to L. This completes option for this term.

Okay, so firstly we want to show that if the limit of X N equals A then the limit of the absolute values of X. N equals the absolute value of A. So we're going to let Episode one Be greater than zero and choose. And in N. Such that for all and greater than or equal to n X n minus A is less than epsilon. Which we can do. Since we know that this limit exists, then we know that for all N greater than or equal to N. So what do we want to show? We want to show that? So let me delete this. We want to show that uh for all and greater than or equal to N, the difference between the absolute value of X. N and the absolute value of A is less than epsilon. So if we can show this then we have this converges to this. But we know that xn minus a. An absolute value like this is less than or equal to x N minus A. And this is by the reverse triangle inequality. The reverse triangle inequality we have that the absolute value of xn minus the absolute value of a. All in absolute value is less than or equal to the absolute value of xn minus A. And we already know that this thing here is less than epsilon. Since that's how we chose our end above. So we have that for all epsilon greater than zero, there exist N N. N. Such that for all N greater than or equal to n X N absolute value minus the absolute value of a is less than epsilon. So we've shown that the limit as X. And the limit as N tends to infinity of the absolute value of X n is equal to the absolute value of a. Now we want to look at the converse. So let's suppose we have a sequence. So that's that part done. Now, I suppose we have a sequence such that the limit as N tends to infinity of the absolute value of X N is equal to the absolute value of A. And we want to ask does the limit as N tends to infinity of xN equal A. And the answer is no. So let's provide a counter example. So let's look at the sequence XN equals -1 to the end. Okay, so let's firstly do this part show that the limit of the absolute value of accent exists. So the absolute value of X n is just one, right? Because this is either one or minus one depending on if N is odd or even. So the absolute value of X n is one for all N. And this converges obviously to one as N tends to infinity. So we have the absolute value of XN converges to one. But the sequence xn doesn't actually converge. So X n is minus one to the N. Which is the sequence minus 11 minus 11 minus 11, et cetera. So -1N doesn't converge. So we find a sequence which converges if you just take absolute values, but the sequence itself doesn't converge. So this is not true this implication.

In the problem we have This integration 0 to a fx. Just relax. General S X dx. Now this is equal to what. So this is equal to let us consider this as you And this as which is. So this is of the form islet that is inverse logarithmic algebraic, programmatic and exponential. So let us assume you as the first function and we s the second Frandsen according to this. So we have this is effects immigration do double ds X dx minus integration. D upon dx off fx integration. General does X. Dicks hold X. So further this can be read in us Effects judas X minus integration. This is F. S. X. Into judas X. Dicks. Further we have to solve this spot. So it becomes F L. S X. Division G D S X dx minus integration. Day upon the excessive every day. Six integration Due to six DX holy X. So this is equal to every six and two jx minus integration if they were last X into gx dicks. So this is the different integration of the spot. So these every sex into JD six. The ecstatic was this one. So further we have the digit of overall integration. So this is overall integration will become now first of all we neglected so we neglect if true and G row as these are equals to zero. So what all integration become integration and you go to a fx G doubled us X dx at equals two F f. A judas A minus. So this is the stamp F or fe into Judith a minus. This is if does uh, g of a minus integration zero to A. If nevertheless X G of x dx. So this is the overall integration. Hence it is the answer to the problem.

We were given the capital omega com api is a finite probability space and that F is the power he said. That is the set of all subsets of omega. First, we'll show that for any random variable X X is F measurable. And then we'll show that the conditional expectation affects given death would equal X itself. Now suppose X is a random variable that is X is a function from capital. And we got to the set of reels. Now for any boils that be over the reals, X inverse B is the set small omega in capital omega. Such that except omega is in B which is clearly a subset of omega. And so must be in the power set S that is X inverse B is an element of F which shows that X is F measurable. Now note that the conditional expectation of X given F is defined as a random variable that satisfies the following two properties number one expectation of X given F is a F measurable random variable and number two or any A N. F. The integral over a of expectation X. Given F. With respect to the probability measure he closed. And to grow over A of X with respect to the probability measure here we have that X itself is an F measurable random variable. Moreover, it trivially satisfies property too. And it's tribute that is trivially true. That integral of way of X with respect to B is equal to the integral over A of X with respect to be So x here satisfies both properties one and 2. And so expectation of X given ETh equals X. Here is important to note that this equality holds almost everywhere. That is, and he set where the conditional expectation X given it does not equal to X has probability measure zero. That's it for this problem.


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