0
$\begingroup$

Let X, Y be two independent Poisson random variables with lambda of X = 1, lambda of Y = 2. Find P(X = 40 | X + Y = 100).

I know P(X|Y) = P(X, Y) / P(Y), and since X and Y are independent P(X, Y) = P(X) P(Y). But P(X) and P(X + Y) are not independent, so how would I go about finding the joint mass function of X and X + Y? I already found the mass function of X + Y.

$\endgroup$

2 Answers 2

Reset to default
0
$\begingroup$

I think you need a division $$P(X=40|X+Y=100)=\dfrac{P(X=40 \text{ and } X+Y=100)}{P(X+Y=100)}=\dfrac{P(X=40 \text{ and } Y=60)}{P(X+Y=100)}=\dfrac{P(X=40)P(Y=60)}{P(X+Y=100)}$$

$\endgroup$
0
$\begingroup$

Hint: If $X = 40$ when we know that $X + Y = 100$, then that means that $Y = 60$. So this is the same probability as $P(X = 40 \text{ and } Y = 60)$. Since these are independent random variables, like you already mentioned, you can rewrite this as a product of simpler probabilities.

Edit: This is slightly wrong, see comments.

$\endgroup$
5
  • $\begingroup$ Hhahahahaha oh wow I'm an idiot. Thanks! Edit: Hmm, so does this mean that P(X = 40 | Y = 60) = P(X = 40), since P(X|Y) = P(X,Y)/P(Y) = P(X)P(Y)/P(Y) = P(X)? (Intuitively this also makes sense since X, Y are independent) $\endgroup$
    – user2197885
    Nov 17, 2013 at 23:27
  • $\begingroup$ That's right! Although, you don't need to use that fact for solving this problem. You should just be able to get away with computing $P(X = 40) \cdot P(Y = 60)$. $\endgroup$
    – Max Sherman
    Nov 17, 2013 at 23:30
  • $\begingroup$ I think you need a division $P(X=40|X+Y=100)=\dfrac{P(X=40 \text{ and } Y=60)}{P(X+Y=100)}=\dfrac{P(X=40)P(Y=60)}{P(X+Y=100)}$ $\endgroup$
    – Henry
    Nov 17, 2013 at 23:31
  • $\begingroup$ Ahh yeah, @Henry is right. I don't think it's as simple as I said it was, unfortunately. $\endgroup$
    – Max Sherman
    Nov 17, 2013 at 23:34
  • $\begingroup$ Yeah, Henry's solution makes sense. Thanks! $\endgroup$
    – user2197885
    Nov 17, 2013 at 23:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .