It is often assumed that the auto traffic that arrives at the intersection during a unit time period has a Poisson distribution with expected value $m$. Assume that the number of cars $X$ that arrive at an intersection from the north in unit time has a Poisson distribution with parameter $m$ and the number $Y$ that arrive from the west in unit time has a Poisson distribution with parameter $\overline{m}$. If $X$ and $Y$ are independent, show that the total number $X + Y$ that arrive at the intersection in unit time has a Poisson distribution with parameter $m+\overline{m}$

#### Solution

Given that $X\sim P(M)$ and $Y\sim P(\overline{m})$.

Moreover $X$ and $Y$ are independently distributed.

Then the moment generating function of $X$ is `$M_{X}(t) =e^{m(e^t-1)}$`

and the moment generating function of $Y$ is `$M_{Y}(t) =e^{\overline{m}(e^t-1)}$`

.

Let $U=X+Y$. Then the moment generating function of $U$ is

`$$ \begin{aligned} M_U(t) &= E(e^{tU}) \\ &= E(e^{t(X+Y)}) \\ &= E(e^{tX} e^{tY}) \\ &= E(e^{tX})\cdot E(e^{tY})\qquad (\because X, Y \text{ are independent })\\ &= M_{X}(t)\cdot M_{Y}(t)\\ &= e^{m(e^t-1)}\cdot e^{\overline{m}(e^t-1)}\\ &= e^{(m+\overline{m})(e^t-1)}. \end{aligned} $$`

which is the moment generating function of Poisson variate with parameter $m+\overline{m}$.

Hence, by uniqueness theorem of moment generating function, $U=X+Y$ follows a Poisson distribution with parameter $m+\overline{m}$.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators