1
$\begingroup$

Random variables are independent and $\mathbb{P}\{X=n\}=\mathbb{P}\{Y=n\}=p, \ n=1,\dots,N$, and $N\in \mathbb{N}$. Find $p$ and $\mathbb{E}(X\mid X+Y=n)$ for $n\in\mathbb{N}$

So the first part:

$\sum^{N}_{n=1} \mathbb{P}\left\{ X=n\right\} =Np=1 \rightarrow p=\frac{1}{N}$

But how do I calculate expected value?

  • 0
    See this: http://www.stat.ufl.edu/~abhisheksaha/sta4321/lect28.pdf2017-02-10

2 Answers 2

1

$P(X=x|X+Y=n)=1/n-1$ ,$x=1,2,3.....n-1$

$E(X|X+Y=n)=\frac{1}{n-1}(1+2+3.....n-1)$

$=n/2$

2

Hint: due to having identical distributions $\mathsf E(X\mid X+Y=n)=\mathsf E(Y\mid Y+X=n)$.