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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.

Karl Kroningfeld
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manayay
  • 297

2 Answers2

1

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

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

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

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

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

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

therefore $Y \sim Poisson(\lambda p)$.

ILikeMath
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-1

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

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

Felix Marin
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