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Problea &: Let X,Y be independlent randot variables On a sample space such that. Poisson( [ ) AId Binomial(4. %) Deternine: () Elex 3Y| (ii) Var(2.1 _ %Y ) (iii...

Question

Problea &: Let X,Y be independlent randot variables On a sample space such that. Poisson( [ ) AId Binomial(4. %) Deternine: () Elex 3Y| (ii) Var(2.1 _ %Y ) (iii) P4 Y =I) (iv) EAAY]

Problea &: Let X,Y be independlent randot variables On a sample space such that. Poisson( [ ) AId Binomial(4. %) Deternine: () Elex 3Y| (ii) Var(2.1 _ %Y ) (iii) P4 Y =I) (iv) EAAY]



Answers

Let $Y$ have a Poisson distribution with mean $\lambda$. Find $E[Y(Y-1)]$ and then use this to show that $V(Y)=\lambda$.

This opposition is given in this question that W is equals two Y. One plus Y. To as Y. One and Y two are independent. So the MGF for W becomes M W. T, which is equals to M. Y. One T. Multiply by M. Y two deep. So therefore it must be so that M W. T divided by M. Y. One T. Which is equals to M. Y. One T. Ask them M. Y. Duty as the and Gfs four W. And Y one is given. So the M. Y. Two becomes he raised by linda Into EU -1. As W has um bison distribution with mean lambda, Where the Y one has a pie, his own distribution with mean λ one, which is less than clammed up. So am Y. Two equals two. E. Raised to par lambda one plus lambda too, in two E. S two party minus one. So this is the and G F. For a pies on distribution with mean λ- λ one. So this is a required answer. Thank you so much.

Yeah, that's probably have been given an equality between the damage distribution. And for Sanders reached now here on this problem, we were told that we have gamma with alpha equals two and beta people don't want and we want to find the probability that why is greater than one. Now notice in the interval that they give us for the Yemen distribution. It goes from lambda to infinity. And certainly from lambda to infinity. We want the probability Y is greater than one. So that just tells us that land is one there is in this girl from I am the to infinitive sort of from one to that. Now matching this up with what we have on the right hand side then that means this probability will be equal to the sun has exposed from zero to alpha minus one. And so it's zero to one of lambda to the X. So one to the arts power times, E to the negative lambda, E to the negative one over ex factory. So again, we just plugged all of the different values that we have been given in the formula that we were given them. And it gives us this summation. Now we can use the table in order to find this or this is easy enough just to find this actual some here and we're going to get the same thing, whether we use the applet or whether we actually use the summation here and finding this gives us 0.7 358 and that's our probability.


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