# Marginal probability derived from joint probability distribution of a poisson process and binomial process

I thought I figured out this problem, but my answer is different than the solutions manual. Hoping someone here can stear me in the correct direction:

Assume the number of automobile collisions that occur on a given stretch of highway per year is a Poisson random variable $X$ with $\mu = 20$. The probability is $p=0.05$ that there will be one or more fatalities in each accident; occurrences of fatalities are independent from one collision to the next. If $Y$ is the number of collisions with one or more fatalities on this stretch of road in one year, find the probability law for $Y$.

So as said, X is poisson, so:

$$p_X(x) = \frac{20^x}{x!}e^{-20}$$

The conditional probability of $Y$ given $X=x$ is a Binomial random variable:

$$p_{Y\mid X}(y\mid x) = \binom{x}{y}p^yq^{x-y}$$

where $Y =$ number of fatal traffic accidents and $x =$ given number of accidents (from poisson distribution)

Therefore the joint probability for $X$ and $Y$ is:

$$p_{X,Y}(x,y)= p_{Y\mid X}\cdot p_X(x) = \binom{x}{y}p^yq^{x-y}\frac{20^x}{x!} e^{-20}$$

to find the marginal probability of $Y$, I sum over $P_{X,Y}$ for all $x$:

\begin{align} p_Y(y) & = \sum_{x=y}^\infty{\binom{x}{y}p^yq^{x-y}\frac{20^x}{x!}e^{-20}} = \frac{(20p)^ye^{-20}}{y!} \sum_{x=y}^\infty {20^{x-y}\frac{q^{x-y}}{(x-y)!}} \\[10pt] & = \frac{(20p)^ye^{-20}}{y!} \sum_{x-y=0}^\infty {20^{x-y}\frac{q^{x-y}}{(x-y)!}} = \frac{(20p)^ye^{-20}}{y!}e^{20q} \end{align}

So:

$$p_Y(y) = \frac{(20p)^ye^{-20p}}{y!}$$

however, the correct answer is apparently $p_Y(y)=\dfrac{e^{-1}}{y!}$

## 1 Answer

\begin{align} & \sum_{x=y}^\infty \frac{20^x e^{-20}}{x!} \cdot \binom x y p^y (1-p)^{x-y} \\[12pt] = {} & \frac{e^{-20} p^y}{y!} \sum_{x=y}^\infty \frac{20^x(1-p)^{x-y}}{(x-y)!} \\[12pt] = {} & \frac{e^{-20} p^y}{y!} \sum_{w=0}^\infty 20^{w+y} \frac{(1-p)^w}{w!} \\[12pt] = {} & \frac{e^{-20} (20p)^y}{y!} \sum_{w=0}^\infty 20^w \frac{(1-p)^w}{w!} \\[12pt] = {} & \frac{e^{-20} (20p)^y}{y!} \cdot e^{20(1-p)} \\[12pt] = {} & \frac{e^{-20p} (20p)^y}{y!}. \end{align} So it's a Poisson distribution with expected value $20p$.

• Please tell as to why should the summing index start from $x=y$. I have difficulty in these concepts – Upstart Feb 21 '16 at 16:30
• You're looking for the probability that the number of accidents involving fatalities is $y$. It is impossible for that number to be $y$ unless there are at least $y$ accidents. $\qquad$ – Michael Hardy Feb 21 '16 at 17:10
• Actually i figured it out just moments after i asked you but still thanku anyways.. – Upstart Feb 21 '16 at 17:23
• So, This is the answer I already got (see my post), however according to the solutions manual, it is not correct. The correct answer is apparently $p_Y(y) = \frac{e^{-1}}{y!}$ . So do you think that the solutions manual is incorrect? – lstbl Feb 22 '16 at 0:40
• @Istbl : They are both correct and they completely agree with each other. You said $\mu=20$ and $p=0.5$. Notice that $20\times0.05 = 1$, i.e. $20p$, where it appears in the answer above, is $1$. So $$\frac{e^{-20p} (20p)^y}{y!} = \frac{e^{-1}}{y!}.$$ – Michael Hardy Feb 22 '16 at 1:50