Find the marginal distribution of $ P\{X=n,Y=m\}=\frac{e^{-14}(7.14)^m(6.86)^{n-m}}{m!(n-m)!}$

Question

Let X denote the number of infants born in one day in a hospital, and Y denote the number of male infants among them. The joint distribution of X and Y is $$ P\{X=n,Y=m\}=\frac{e^{-14}(7.14)^m(6.86)^{n-m}}{m!(n-m)!}\\ n=0,1,2,...;\space m=0,1,2,...,n. $$

The question is to find the marginal distribution.

I know: $$P\{X=n\}=\displaystyle \sum_{m=0}^{n} \frac{e^{-14}(7.14)^m(6.86)^{n-m}}{m!(n-m)!}, P\{Y=m\}=\displaystyle \sum_{n=m}^{\infty} \frac{e^{-14}(7.14)^m(6.86)^{n-m}}{m!(n-m)!}$$.

But I don't know how to calculate or simplify the two formulas.

Answer 1

Just to make it a little bit easier for me to write the solution let me solve for a more "general" case where the joint distribution of $X$ and $Y$ is $$ P\left\{X=n,Y=m\right\}=\frac{\lambda^{-n}e^{-\lambda}p^{m}\left(1-p\right)^{n-m}}{m!(n-m)!} $$ In your example $\lambda=14$ and $p=7.14/14$. Let's solve for $X$: $$ P\left\{X=n\right\}=\sum_{m=0}^{n}\frac{\lambda^{-n}e^{-\lambda}p^{m}\left(1-p\right)^{n-m}}{m!(n-m)!}=\frac{\lambda^{n}e^{-\lambda}}{n!}\sum_{m=0}^{n}\frac{n!}{m!(n-m)!}p^{m}\left(1-p\right)^{n-m} $$ Using Newton's Binomial Theorem the series is equal to $\left(p+1-p\right)^{n}=1$ and thus $X$ has Poisson Distribution with parameter $\lambda$.

Now let's solve for $Y$ (you made a tiny mistake when defining the inferior limit of the series, it starts from $m$): $$ P\left\{Y=m\right\}=\sum_{n=m}^{\infty}=\frac{\lambda^{-n}e^{-\lambda}p^{m}\left(1-p\right)^{n-m}}{m!(n-m)!}=\frac{\lambda^{-m}e^{-\lambda}}{m!}\sum_{m=n}^{\infty}\frac{\left(\lambda(1-p)\right)^{n-m}}{(n-m)!} $$ Using a change of variable $v=n-m$ you get that the series is the Maclaurin series of $e^{\lambda(1-p)}$. Hence, $Y$ has Poisson Dsitribution with parameter $\lambda p$.

To conclude, I would like to add some comments regarding this problem. You can interpret this problem in a "Bayesian way" as follows: let $X$ be the number of babies born in one day, then $X\sim Po(\lambda)$. The probability that a baby is a boy is $p$, so let $Y$ be the number of boys born in a day, then if $X$ babies where born we have that $Y|X\sim Bin(X,p)$ (you can divide the joint by the marginal and check this).

Answer 2

From the Binomial Theorem, $$P(X=n)=\sum_{m=0}^{n}\frac{e^{-14}(7.14)^m(6.86)^{n-m}}{m!(n-m)!}=\frac{e^{-14}}{n!}\sum_{m=0}^{n}{n \choose m}(7.14)^m(6.86)^{n-m}=\frac{e^{-14}}{n!}(7.14+6.86)^n$$ Moreover, the lower index on your sum corresponding to $P(Y=m)$ should begin at $n=m$. If you shift the index you get $$P(Y=m)=\sum_{n=m}^{\infty}\frac{e^{-14}(7.14)^m(6.86)^{n-m}}{m!(n-m)!}=\frac{e^{-14}}{m!}\cdot \bigg(\frac{7.14}{6.86}\bigg)^m\cdot\sum_{n=m}^{\infty}\frac{(6.86)^n}{(n-m)!}=\frac{e^{-14}(7.14)^m}{m!}\cdot\sum_{n=0}^{\infty}\frac{(6.86)^n}{n!}=\frac{e^{-14+6.86}(7.14)^m}{m!}$$