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I thought I would be able to do these problems, all I can find are proofs that if random variables are independent that the product of distributions is the joint distribution of the two. I cannot actually compute it. So to show that, say, two discrete random variables $X,Y$ taking values in $\mathbb{Z}$ are independent you need to show that

$\mathbb{P}(X=x,Y=y)=\mathbb{P}(X=x)\mathbb{P}(Y=y)$ for all $x,y\in\mathbb{Z}$

For example, if $X\sim\text{Pois}(\nu p)$ and $Y\sim\text{Pois}(\nu)$, ($X,Y$ independent) I would like to show that $X$ and $Y-X$ are independent random variables.

I thought the best way to go was:

$$\mathbb{P}(X=x,Y-X=y)=\mathbb{P}(X=x,Y=x+y)=\frac{(\nu p)^x}{x!}\exp(-\nu p)\frac{\nu^{x+y}}{(x+y)!}\exp(-\nu)$$

Is this because $X$ and $Y$ are independent, and you have reformulated it in terms of independent random variables? Is this the general method for problems with differences of random variables like $X-Y$ and $X$ with $X,Y$ independent?

Edit: Obviously this would be convenient because then $$\mathbb{P}(X=x)=\frac{\nu^x}{x!}\exp(-\nu)$$

and then, the rest looks like:

$$\mathbb{P}(X-Y=y)=\frac{\nu^{x+y} p^x}{(x+y)!}\exp(-\nu p)$$

Is this the part where I recognise the distribution and conclude it is independent or where I deduce what the distribution is and that it is independent some other way?

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I think he means Bernoulli. What's a Poi distribution? Oriental fish perhaps? As distinct from the French variety. – wolfies May 1 '13 at 12:34
There was a mistake in my original post I have edited it. I will change $\text{Poi}$ to $\text{Poisson}$ if it is confusing – shilov May 1 '13 at 12:37
changed it to $X\sim \text{Pois}(\nu)$ – shilov May 1 '13 at 12:55
Why do you think they would be independent? Surely $\mathbb E(Y-X |X=x) = \mathbb E(Y) - x$ depends on $x$. So they can't be independent. – Tim May 1 '13 at 17:56
That doesn't show that they can't be independent – shilov May 1 '13 at 19:51

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