poisson random variable distribution using probability Let $X$ be the number of emails that a company receives in a day. Assume that $X$ is a Poisson random variable with parameter $\lambda$. The company classifies each email as spam or not spam. The probability that a single email is spam is $p$. Let $Y$ be the number of spam emails that the company receives in a day. Assume that for any two emails, whether or nor they are spam are events.
  (a) Find $E[Y]$.
  (b) Show that $Y$ is also a Poisson random variable and find its parameter (in terms of $\lambda$ and $p$). Hint: Use the partition theorem: $$P(Y=k)=\sum\limits_{n=k}^{\infty}{P(Y=k|X=n)P(X=n)}.$$


I have tried it out, but I am finding it difficult to post them in proper format.
Please help me with the solution.
Firstly we will prove $(b)$, then $(a)$ can be deduced easily. We will show that $Y \sim Poisson(\lambda p)$ using the fact that the Moment-generating function of $Y$ will be the Poisson one.

We have $X \sim Poisson(\lambda)$ and $\displaystyle M_X(t)=e^{\lambda\left(e^t - 1\right)}.$
Now 

$$\begin{align}
M_Y(t)&amp;=E[e^{it}]\\
&amp;=\sum\limits_{i=0}^{\infty}{e^{it}P\left( Y=i \right)}\\
&amp;=\sum\limits_{i=0}^{\infty}{e^{it}\left( \sum\limits_{j=i}^{\infty}{P(  Y=i|X=j)P\left(X=j\right)}\right)}
\end{align}$$

but $\displaystyle P(Y=i|X=j)=\binom{j}{i}p^iq^{j-i}=\frac{j!}{(j-i)!i!}p^iq^{j-i}$, i.e. binomial, because the events asociated with probability $p$ are independet and $\displaystyle P(X=j)=\frac{e^{-\lambda} \lambda^j}{j!}$. So

$$\begin{align}
M_Y(t)
&amp;=\sum\limits_{i=0}^{\infty}{e^{it}\left( \sum\limits_{j=i}^{\infty}{\frac{j!}{(j-i)!i!}p^iq^{j-i}\frac{e^{-\lambda} \lambda^j}{j!}}\right)} \\
&amp;=e^{-\lambda}\sum\limits_{i=0}^{\infty}{e^{it}\frac{1}{i!}\frac{p^i}{q^i}\sum\limits_{j=i}^{\infty}{\frac{q^{j}\lambda^j}{(j-i)!}}} \\
&amp;=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\frac{e^{t}p}{q}\right)^i\sum\limits_{j=i}^{\infty}{\frac{(q\lambda)^{j}}{(j-i)!}}} \\
&amp;=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\frac{e^{t}p}{q} \right)^i\sum\limits_{j=i}^{\infty}{\frac{(q\lambda)^i(q\lambda)^{j}}{(q\lambda)^i(j-i)!}}} \\
&amp;=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\frac{e^{t}p}{q} \right)^i(q\lambda)^i\sum\limits_{j=i}^{\infty}{\frac{(q\lambda)^{j-i}}{(j-i)!}}} \\
&amp;=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\lambda e^{t}p \right)^i\sum\limits_{j=0}^{\infty}{\frac{(q\lambda)^{j}}{j!}}} \\
\end{align}$$

but we know $\displaystyle \sum\limits_{k=0}^{\infty}{\frac{x^{k}}{k!}}=e^x,$ hence

$$\begin{align}
M_Y(t)
&amp;=e^{-\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\lambda e^{t}p \right)^i e^{q\lambda}} 
=e^{-\lambda}e^{q\lambda}\sum\limits_{i=0}^{\infty}{\frac{1}{i!}\left(\lambda e^{t}p \right)^i }
=e^{-\lambda(1-q)}e^{\lambda e^{t}p}=e^{\lambda p\left( e^t -1\right)},
\end{align}$$

therefore $Y \sim Poisson(\lambda p)$.
$\large p \gtrsim 0\,,\quad X \gg 1$.
\begin{align}
{X \choose Y}\,p^{Y}\,\left(1 - p\right)^{X - Y}
&amp;=
{X\left(X - 1\right)\ldots\left(X - Y + 1\right) \over Y!}\,
p^{Y}\exp\left(\vphantom{\LARGE A}\left[\,X - Y\,\right]\,\ln\left(1 - p\right)\right)
\end{align}
Since
$$
\left\lbrace%
\begin{array}{rcl}
X\left(X - 1\right)\ldots\left(X - Y + 1\right)
&amp; = &amp;
X^{Y}\left(1 - {1 \over X}\right)\ldots\left(1 - {Y - 1 \over X}\right)
\approx
X^{Y}
\\
\exp\left(\vphantom{\LARGE A}\left[X - Y\right]\,\ln\left(1 - p\right)\right)
&amp; \approx &amp; 
\exp\left(\vphantom{\LARGE A}X\,\left[-p\right]\right)
= {\rm e}^{-Xp}
\end{array}\right.
$$

we get
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
{X \choose Y}\,p^{Y}\,\left(1 - p\right)^{X - Y}
\color{#000000}{\ \approx\ }
{X^{Y} \over Y!}\,p^{Y}{\rm e}^{-Xp}
\color{#000000}{\ =\ }
{\left(Xp\right)^{Y}{\rm e}^{-Xp} \over Y!}\quad}
\\ \\ \hline
\end{array}
$$