0

We have $X$ ~Poisson(a) and $Y$~Poisson(b), where $X$ and $Y$ are independent random variables. We are asked to calculate $P( X = k | X + Y = n)$ for $k$ between $[0,n]$. I know $(X + Y)$ is Poisson $(a + b)$, which is used in book's solution; however I use a different method.

I think that answer of this probability is equal to $P(Y = n-k)$; however answer in the book is different. If we are given that $X+Y = n$, the only way that $X = k$ is $Y = n-k$. Could you please explain what am I missing ? Thanks in advance.

p_square
  • 1,007
  • 1
  • 6
  • 23
levitatmas
  • 7

1 Answers1

2

Essentially, you are aguing, that if $X+Y=n$, then $X=k$ if and only if $Y=n-k$. This is correct, but it only tells us that the conditional probabilities of the events $X=k$ and $Y=n-k$ are the same, when we condition on $X+Y=n$. Written mathematically this means that $$\mathbb{P}(X=k \: | \: X+Y = n) = \mathbb{P}(Y=n-k \: | \: X+Y=n),$$ which is NOT the same as claiming that $$\mathbb{P}(X=k \: | \: X+Y = n) = \mathbb{P}(Y=n-k).$$

Correct approach:

Use the definition of conditional probability: $$\mathbb{P}(A|B) = \frac{\mathbb{P}(A,B)}{\mathbb{P}(B)}.$$ This gives us that \begin{align*} \mathbb{P}(X=k \:|\: X+Y=n) &= \frac{\mathbb{P}(X=k,X+Y=n)}{\mathbb{P}(X+Y = n)} \\ &= \frac{\mathbb{P}(X=k,Y=n-k)}{\mathbb{P}(X+Y = n)} \\ &= \frac{\mathbb{P}(X=k)\mathbb{P}(Y=n-k)}{\mathbb{P}(X+Y = n)} \end{align*} Insert the correct PMF's and you are done.

Leander Tilsted Kristensen
  • 5,488