I had a probability question I was hoping for some help on! First, the question:
Let $Y_1$ and $Y_2$ be independent Poisson random variables with means $λ_1$ and $λ_2$, respectively.
a) Find the probability density function of $Y_1 + Y_2$.
b)Find the conditional probability density function of $Y_1$, given that $Y_1 + Y_2 = m$
By doing the method of using moment generating functions, I was able to find a) such that:
$M_{y1} (t) = e^{λ(e^t -1)}$
$U=Y_1+Y_2$
$M_u(t) = M_{y1}(t) * M_{y2}(t)$
$M_u(t) = e^{λ_1(e^t -1)} * e^{λ_2(e^t -1)}$ = $e^{(λ_1+λ_2)(e^t-1)}$
Thus, $U= Y_1 + Y_2$ follows Poisson with mean ($λ_1 + λ_2$); Thus giving the probability density function $p(y) = \dfrac{λ^ye^{-λ}}{y!}$
However, b) is much trickier. I know that the conditional function is
$P(Y_1|Y_1 + Y_2 = m) = \dfrac{P(Y1,Y1+Y2 = m)}{P(Y1+Y2 = m)}$ = $\dfrac{P(Y1)*P(Y1+Y2 = m)}{P(Y1+Y2 = m)}$
because of the independence of the variables. However, I'm not sure how to solve this because if it is so, wouldn't the $P(Y_1+Y_2 = m)$ 's just cancel? Would someone be able to help me? Thank you so much in advanced, I really appreciate it!
Best Answer
What you write in your solution to b) does not make sense ($P(Y_1)$ is undefined, as $Y_1$ is a random variable, not an event). The probability density function of $(Y_1\mid Y_1+Y_2=m)$ is by definition $$ P(Y_1=k\mid Y_1+Y_2=m),\quad k\ge0. $$ This is zero if $k>m$ and if $k\le m$, by Bayes' theorem, \begin{align*} P(Y_1=k\mid Y_1+Y_2=m) =\frac{P(Y_1=k,Y_1+Y_2=m)}{P(Y_1+Y_2=m)} &=\frac{P(Y_1=k,Y_2=m-k)}{P(Y_1+Y_2=m)}\\ &=\frac{P(Y_1=k)\,P(Y_2=m-k)}{P(Y_1+Y_2=m)}, \end{align*} by independence of $Y_1$ and $Y_2$. At this point, you should be able to finish the computation since you know the pdf of $Y_1$, $Y_2$ and $Y_1+Y_2$.