# Conditional independence problem for poisson random variables

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!

• Maybe use that $X=x$ is the same as $V+W = x$ which is the same as $W = x-V$, do the same for $Y$. Oct 16, 2019 at 10:34
• @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?
– Wyuw
Oct 16, 2019 at 10:43
• From here I would just state that W and Z are independent so the P_{Z,W}(.) factorize as shown by @gunes Oct 16, 2019 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}

• 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.
– Wyuw
Oct 16, 2019 at 10:56
• It is the definition of conditional probability, i.e. $$P(A|B)=\frac{P(A,B)}{P(B)}$$ in its simplest form. Oct 16, 2019 at 10:57
• Oh ok. But what was used for the separation $P(W=w,Z=z,V=v)=P(W=w)P(Z=z)P(V=v)$?
– Wyuw
Oct 16, 2019 at 11:01
• That comes from the independence assumption stated in the question. Oct 16, 2019 at 11:02
• Ahh, this all makes sense to me now! Thank you very much!!!
– Wyuw
Oct 16, 2019 at 11:03