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Mx(t)EXVar Xprnf / pdf Px(I;p) =p(1 -P)" ~* I =0,1 Px(r;n,P) = ()p(1 - p)"-* , 1 =01. Px(r;p) (1 -P)-'p, I=l2.- px(r;p,r) = (-H)( - p)r-p , Ienn+l Px...

Question

Mx(t)EXVar Xprnf / pdf Px(I;p) =p(1 -P)" ~* I =0,1 Px(r;n,P) = ()p(1 - p)"-* , 1 =01. Px(r;p) (1 -P)-'p, I=l2.- px(r;p,r) = (-H)( - p)r-p , Ienn+l Px(I;^) =e-^A/rl I=ul- Px(I; N,M,K) = ()(*-")() I=o,lK px(5;K) = % T=- px(I;fhIo) T=T fx(r;/,o") = {d exp ~C <I < 00 Jx(c;0,8) = Noysr"-1 exp 0 < I < 00 fx(r;0,8) = Fatosr"-'(I-r)-'0< 1 < 1pe' + (1 -P) Ipe" + (1 ~ p)l"p(1 - p) np(1 (1 ~ pp-? ~(1 - pp--7 4 MlE MF) H-N 4 (KH-W

Mx(t) EX Var X prnf / pdf Px(I;p) =p(1 -P)" ~* I =0,1 Px(r;n,P) = ()p(1 - p)"-* , 1 =01. Px(r;p) (1 -P)-'p, I=l2.- px(r;p,r) = (-H)( - p)r-p , Ienn+l Px(I;^) =e-^A/rl I=ul- Px(I; N,M,K) = ()(*-")() I=o,lK px(5;K) = % T=- px(I;fhIo) T=T fx(r;/,o") = {d exp ~C <I < 00 Jx(c;0,8) = Noysr"-1 exp 0 < I < 00 fx(r;0,8) = Fatosr"-'(I-r)-'0< 1 < 1 pe' + (1 -P) Ipe" + (1 ~ p)l" p(1 - p) np(1 (1 ~ pp-? ~(1 - pp-- 7 4 MlE MF) H-N 4 (KH-W 5= En 1)? 646'-1) icotplicadisimol #EK,e Ee" euton (1 - Bt)-o 48' E#(K#) 7+pytaT8+1)



Answers

Let $X \sim \operatorname{Unif}[A, B],$ so its pdf is $f(x)=1 /(B-A), A \leq x \leq B, f(x)=0$ otherwise. Show that the moment generating function of $X$ is
$$M_{X}(t)=\left\{\begin{array}{cc}{\frac{e^{B t}-e^{A t}}{(B-A) t}} & {t \neq 0} \\ {1} & {t=0}\end{array}\right.$$

Hey, guys. So it's clear for here. So the pdf x is, um, given to us and we can calculate the expected value your steps as into girl, from Dana to Infinity. X Times, Katy Times data to K power over Becks two the Cape plus one. Jax. And we got K Times data overcame minus one for part B. If X is equal to one, then our analysis will be the same so marked in the previous part here. So we're gonna pick up there and just substitute K is equal to one. Then we got you have fax is equal to so you, uh, Infinity Kenny times They'd have to keep our over x to the k t X just equal to integral off data to infinity data over X t X who just equal to Data times well and Max from data to infinity, which is legal too. The times l end of infinity minus fell. And if they do, which is under find for a part. See, weaning is calculate me of next square, which we got into girl from data to K that square times K Times data to the Cape power over next to the K plus one de ETS, which is equal to K times Dina to the K. What for? Came minus two cons won over thehe to the king minus two. Power in this equals Okay, times they'd a square. Okay, you times they does square overcame on this too for a party. Oh, no. Sorry. Excuse me. We can write V of X a cz b to the X square minus need to that X square. And this is equal to Kenny times. They need a square. Overcame minus two minus k times they'd up over Canaan minus one squared, which is equal to K Times Data square over a minus two. Someone's came, Linus. One square per party. If K is equal to two, then vfx will be infinite. Been since the interval and expression of, um e to the x square will evaluate to the infinity. So K equals two. And be a Lex. Well, you motion sense p of x square. Well, if all you to infinity her part being, um So we're party. We can conclude that to ensure that e of extra and power is finite is that king has bigger than in

In this exercise we refer back to the pdf given an exercise 29 which I have shown here. And we also make use of the following information. So we define a function l sub x t, which is the natural logarithms of the MGF then the first derivative of L sub x At T equals zero is equal to the expected value of X and the second derivative is equal to the variance of X and we begin in part A by finding the moment generating function for X and using the MGF to determine the expected value and the variance of X. By definition, the moment generating function is the expectation of eating exponents Tee times x. I'm integrating starting at zero because our function is only defined for X is greater than or equal to zero. Now we evaluate this over zero to infinity. Now note that the integral Onley converges if t is less than four. If he is greater than four, then it blows up as X approaches infinity In order to find the expected value, we first find the first derivative of the moment Generating function gives 4/4 minus t squared so the expected value is the moment generating function derivative when he is set to zero, which is 1/4. To find the variants weaken, begin by finding the second derivative of the moment generating function. That's 8/4 minus t to the explain it. Three. The expectation of X squared is equal to the second derivative of the moment generating function when t is set to zero and this gives 1/8. So the variance of X using the very in shortcut formula is the expectation of X squared minus the expectation of X squared. This comes out to 1/16, which means that our standard deviation is 1/4. So we have found the answers to the expectation and standard deviation using the moment generating function and now for part B, were asked once again to find the expectation and the variance of X. But to do so using the function. Elsa, Becks, this function is a natural algorithm of the moment generating function. So it's the natural algorithm of 4/4 minus t. Now, in order to find the expectation, we first find the first derivative of Elsa Becks. So this part here is just one over this part. So the first derivative comes out to 1/4 minus t. Now we confined the expected value of X, which is equal to the first derivative sales of X At T equals zero gives us 1/4, which matches? Our answer used. Our answer obtained using the moment generating function. To find the variance of X, we find the second derivative of L sub X. This is simply 1/4 minus t in brackets squared. Now the variance as the second derivative of Elsa, Becks with tea set equal to zero, gives us one of her 16, and that gives us a standard deviation of 1/4, which again is an agreement.


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