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$X_1,...,X_n$ are independent, and $X_i\sim Poisson(i)$, the question ask for the $pdf$ of $\overline X$, and the $pmf$ of $X_{(n)}$.

Here is what I tried:

1. Since they are $Poisson$ distribution, so from $mgf$ of $Poisson$ distribution, $M_{X_{i}}(t)=e^{i(e^t-1)}$. Then since $X_i$ are indep, and let $Y=\sum{X_i}$, so, $M_Y(t)=E(e^{tY})=E(e^{t\sum{X_i}})=E(e^{tX_1} \times \cdots\times e^{t\sum{X_i}})=E(e^{tX_1})\times\cdots\times E(e^{tX_n})=\Pi_{i=1}^n E(e^{tX_i})=\Pi_{i=1}^n e^{i(e^t-1)}=e^{\frac{n(n+1)}{2}(e^t-1)}$

$Y=\sum X_i\sim Poisson(\frac{n(n+1)}{2})$, then $\overline X=\frac{Y}{n}$, so,

$f_{\overline X}(x)=f_{Y}(nx)|\frac{d(nx)}{x}|=\frac{e^{-\frac{n(n+1)}{2}}}{(nx)!}(\frac{n(n+1)}{2})^{nx}n$. I'm not sure whether I was wrong or not on this part, because the transformation formula I found for the discrete RV and continuous RV are different, and here, the transformation I used is from a discrete RV to a continuous RV, I'm got a little confused here whether I should add $n$ as the end of the $pdf$, or may be I was totally wrong.

2. The $pmf$ of $X_{(n)}$ is $P(X_{(n)}=x)=P(X_{(n)}\le x)-P(X_{(n)}\le x-1)=P(X_{1}\le x,\cdots,X_{n}\le x)-P(X_{1}\le x-1,\cdots,X_{n}\le x-1)=\Pi_{i=1}^n P(X_{i}\le x)-\Pi_{i=1}^n P(X_{i}\le x-1)=\Pi_{i=1}^n\sum_{t=0}^{x}\frac{e^{-i}}{t!}i^t-\Pi_{i=1}^n\sum_{t=0}^{x-1}\frac{e^{-i}}{t!}i^t$.

so here, can I switch the product and the summation? Moreover, through this way, I feel I'm already messed up with the calculation.

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the transformation I used is from a discrete RV to a continuous RV...

A random variable is discrete if it only can take on a discrete number of values. Since $\bar{X}$ is a finite sum of discrete random variables so it is also discrete. So you should not have the factor of n. In this case you can see \bar{X} is discrete because it can now take on number m/n where m is an integer (because it is a sum of the values of other Poissonian distributions that are integer valued), these are are still fractions which is a discrete set.

With regards to the pmf of $X_{(n)}$. You cannot switch the sum, for example $(a+b)(c+d) \ne ac +db$ which would be equivalent to switching products and sums. Also you did not define $X_{(n)}$, but I'm assuming from your calculation it is $\max_i X_i$.

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  • $\begingroup$ Thank you, I did work it out with your help! Your idea is totally correct! $\endgroup$ – user456649 Jun 26 '17 at 22:13

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