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I have solved the a and b part and one half of the c part. I am getting $$P(X=x|Y=y) ~ Poisson(\lambda (1-p))$$ And when i try to get the expectation its not right .i get it as $$\lambda(1-p)$$ but the solution manual says its $$y+\lambda(1-p)$$..please someboday explain this concept..
Observe that \begin{align} p_{X,Y}(n,m) &= \frac{\lambda^n e^{-\lambda} p^m(1-p)^{n-m}}{m!(n-m)!}\\ &= \left(\frac{\lambda^n e^{-\lambda}}{n!}\right)\binom nm p^m(1-p)^{n-m}, \end{align} so $Y\mid X\sim\operatorname{Bin}(X,p)$, that is, the conditional pmf of $Y$ given $X=n$ is $$p_{Y\mid X=n}(m\mid n) = \binom nm p^m(1-p)^{n-m}.$$
To compute the marginal pmf of $Y$, we can sum the joint pmf over all values for $X$:
\begin{align} p_Y(m)&=\mathbb P(Y=m)\\ &= \sum_{n=m}^\infty \mathbb P(X=n,Y=m)\\ &= \sum_{n=m}^\infty \frac{\lambda^n e^{-\lambda} p^m(1-p)^{n-m}}{m!(n-m)!}\\ &= \frac{e^{-\lambda}p^m}{m!}\sum_{n=m}^\infty \frac{\lambda^n (1-p)^{n-m}}{(n-m)!}\\ &= \frac{e^{-\lambda}\lambda ^m p^m}{m!}\sum_{n=0}^\infty \frac{\lambda^n (1-p)^n}{n!}\\ &= \left(\frac{e^{-\lambda}(\lambda p)^m}{m!}\right)e^{\lambda(1-p)}\\ &= \frac{e^{-\lambda p}(\lambda p)^m}{m!}, \end{align} and so $Y\sim\operatorname{Pois}(\lambda p)$. The conditional pmf of $X$ given $Y$ is then \begin{align} p_{X\mid Y=m}(n\mid m) &= \frac{p_{X,Y}(n,m)}{p_Y(m)}\\ &= \left(\frac{\lambda^n e^{-\lambda} p^m(1-p)^{n-m}}{m!(n-m)!}\right)\left(\frac{m!}{e^{-\lambda p}(\lambda p)^m} \right)\\ &= \frac{(\lambda(1-p))^{n-m}e^{-\lambda(1-p)}}{(n-m)!}. \end{align}
To compute the marginal pmf of $X$, we can sum the joint pmf over all values for $X$:
\begin{align} \mathbb P(X=n) &= \sum_{m=0}^n \mathbb P(X=n\mid Y=m)\\ &= \sum_{m=0}^n \left(\frac{\lambda^n e^{-\lambda}}{n!}\right)\binom nm p^m(1-p)^{n-m}\\ &= \left(\frac{\lambda^n e^{-\lambda}}{n!}\right)\sum_{m=0}^n \binom nm p^m(1-p)^{n-m}\\ &= \frac{\lambda^n e^{-\lambda}}{n!}, \end{align} and so $X\sim\operatorname{Pois}(\lambda)$.
The conditional expectations can be computed directly from the conditional pmfs:
\begin{align} \mathbb E[X\mid Y=m] &= \sum_{n=m}^\infty np_{X\mid Y=m}(n\mid m)\\ &= \sum_{n=m}^\infty \frac{n(\lambda(1-p))^{n-m}e^{-\lambda(1-p)}}{(n-m)!}\\ &= \sum_{n=0}^\infty \frac{(n+m)e^{-\lambda(1-p)}(\lambda(1-p))^n}{n!}\\ &= \sum_{n=0}^\infty n\frac{e^{-\lambda(1-p)}(\lambda(1-p))^n}{n!} + m\sum_{n=0}^\infty \frac{(n+m)e^{-\lambda(1-p)}(\lambda(1-p))^n}{n!}\\ &= \lambda + m, \end{align} and
\begin{align} \mathbb E[Y\mid X=n] &= \sum_{m=0}^n m\binom nm p^m(1-p)^{n-m}\\ &= np. \end{align}