$X$ and $Y$ are linearly independent random variables with geometric distribution. They have the same $p$ parameter. How can I calculate the following expressions:
a) $P(X=k|X<Y)$=?
b) $P(X=k|X=Y)$=?
c) $\mathbb{E}(X\mid X+Y)$=?
For b) I got: $(1-p)^{2k}[(1-p)^{-2}+1]$
I made a matrix, and its elements was $P(X=i;Y=j)$ ... I took the sum of the diagonal elements and I divided $P(X=k;Y=k)=p^{2}(1-p)^{2k-2}$ with this sum. Is it a correct solution?
a) was harder...
$P(X=k\mid X<Y)=\frac{P(X=k;X<Y)}{P(X<Y)}$
${P(X<Y)}$ is the sum of the elements in the upper triangle matrix, because this matrix is symmetric. The "probability" of the upper triangle matrix is
$\left(1-(sum\:of\:the\:diagonal\:elements)\right)/2$
$P(X=k|X<Y)$ is $p(1-p)^{k-1}\sum_{j=k+1}^{\infty}(1-p)^{j-1}p=p(1-p)^{k-1}\cdot\left(\sum_{j=1}^{\infty}(1-p)^{j-1}p-\sum_{j=1}^{k}(1-p)^{j-1}p\right)$
I divided this result with $P(X<Y)$ (calculated above) and I got the following disgusting expression:
$2\frac{(1-p)^{2k-1}p}{1-\frac{p^{2}}{1-(1-p)^{2}}}$
Is it correct? I'm working on task c) but I would appreciate your help :DD
For a), we have \begin{align} \mathbb P(X=k\mid X<Y) &= \frac{\mathbb P(X=k,X<Y)}{\mathbb P(X<Y)}\\ &= \frac{(1-p)^{k-1}p\sum_{i=k+1}^\infty (1-p)^{i-1}p}{\sum_{i=1}^\infty \sum_{j=1}^{i-1} (1-p)^{i-1}p(1-p)^{j-1}p}\\ &= \frac{p (1-p)^{2 k-1}}{\frac{1-p}{2-p}}\\ &= p(2-p) (1-p)^{2 (k-1)}. \end{align}
For b), we have \begin{align} \mathbb P(X=k\mid X=Y) &= \frac{\mathbb P(X=k,X=Y)}{\mathbb P(X=Y)}\\ &= \frac{(1-p)^{k-1}p(1-p)^{k-1}p}{\sum_{i=1}^\infty (1-p)^ip(1-p)^ip}\\ &= p(2-p) (1-p)^{2 (k-1)}. \end{align}
For c) we have for each positive integer $n$
\begin{align} \mathbb E[X\mid X+Y=n] &= \frac{\mathbb E\left[X\mathsf 1_{\{X+Y=n\}}\right]}{\mathbb P(X+Y=n)}\\ &= \frac{\mathbb E\left[X\mathsf 1_{\{X+Y=n\}}\right]}{1 - 2\left(\frac{1-p}{2-p}\right)}. \end{align}
I am not sure how to evaluate $\mathbb E[X\mathsf 1_{\{X=Y\}}]$, however.