I have this problem:

Let $X = V + W$ and $Y = V + Z$ where $V, W, Z$ are independent Pois($\lambda$) random variables.

I found that $Cov(X, Y) = Var(V) = \lambda$

It now asks to find whether $X$ and $Y$ are conditionally independent given $V$.

I am now trying to find if $X$ and $Y$ are conditionally independent given $V$.

So I start with $P(X = x, Y = y | V = v) =$

Now, I know that, for conditional independence, I have to show that the joint probability mass function factors into the product of its marginal probability mass functions.

But where do I go from here? This is where I have been stuck.

Please show me how this is done. Thank you very much for your help!

  • $\begingroup$ Maybe use that $X=x$ is the same as $V+W = x$ which is the same as $W = x-V$, do the same for $Y$. $\endgroup$ – Jesper for President Oct 16 '19 at 10:34
  • $\begingroup$ @JesperHybel $P(X = x, Y = y | V = v) = P(W = x - v, Z = y - v | V = v)$ Hmm, I think it is $P(X = x, Y = y | V = v) = P(W = x - v, Z = y - v | V = v) = P(W = x - v, Z = y - v)$ because we already are given $v$, so the $| V = v$ part is redundant, right? $\endgroup$ – Wyuw Oct 16 '19 at 10:43
  • $\begingroup$ From here I would just state that W and Z are independent so the P_{Z,W}(.) factorize as shown by @gunes $\endgroup$ – Jesper for President Oct 16 '19 at 10:55

Intuitively, if $V$ is known, the only random component in $X$ will be $W$ and, in $Y$, it will be $Z$. Since the two are independent, so are $X$ and $Y$. More mechanically, you can start from your and @Jesper's comments: $$\begin{align}P(W=w,Z=z|V=v)&=\frac{P(W=w,Z=z,V=v)}{P(V=v)}\\&=\frac{P(W=w)P(Z=z)P(V=v)}{P(V=v)}\\&=P(W=w)P(Z=z)\end{align}$$

  • $\begingroup$ Ahh. At least I was on the right track. What rule is used to get $P(W=w,Z=z|V=v)=\frac{P(W=w,Z=z,V=v)}{P(V=v)}$? Some form of Bayes' rule? I am unsure of what rule was used to get each of the equality. Sorry, I am just learning this. $\endgroup$ – Wyuw Oct 16 '19 at 10:56
  • $\begingroup$ It is the definition of conditional probability, i.e. $$P(A|B)=\frac{P(A,B)}{P(B)}$$ in its simplest form. $\endgroup$ – gunes Oct 16 '19 at 10:57
  • $\begingroup$ Oh ok. But what was used for the separation $P(W=w,Z=z,V=v)=P(W=w)P(Z=z)P(V=v)$? $\endgroup$ – Wyuw Oct 16 '19 at 11:01
  • $\begingroup$ That comes from the independence assumption stated in the question. $\endgroup$ – gunes Oct 16 '19 at 11:02
  • $\begingroup$ Ahh, this all makes sense to me now! Thank you very much!!! $\endgroup$ – Wyuw Oct 16 '19 at 11:03

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