1

I have a question which says two discrete random variables $X, Y$ have joint probability function $$ f(x,y) = {n\choose {x, y, n-x-y}}p^xq^y(1-p-q)^{n-x-y} $$ Then the question asks for the distribution of $T = X+Y$. It is easy to see that $T$ counts the number of occurrences of $X$ or $Y$, so $T\sim\text{Binomial}(n, p+q)$. However, from $f(x, y)$ we can expect the marginal probability functions $f_X$ and $f_Y$ to behave like binomial distributions as well, since for $f_X$ we are simply counting $X$ and "not $X$" (analogously for $Y$), so $X\sim\text{Binomial}(n,p)$ and $Y\sim\text{Binomial}(n,q)$.

However, this question seems to make much more a fuss about the sum of binomial distributions not having an easy formula. So who is correct? Is the sum of binomial distributions with same $n$ but different probabilities distributed like a binomial distribution with $n$ trials and probability = to the sum of probabilities?${}^1$ Does it work in this case because my $X$ and $Y$ are dependent?

$1$ - I'm inclined not to believe this because it would imply that if $X\sim\text{Binomial}(n,p)$, then $kX\sim\text{Binomial}(n,kp)$, and $kp$ is unbounded, so it can be greater than $1$, which doesn't make any sense.

user3002473
  • 9,245

1 Answers1

1

In general if $X$ and $Y$ have binomial distributions with the same $n$, their sum need not have an easy formula, so the fussy question is right. Depending on the joint distribution of $X$ and $Y$, the distribution of $X+Y$ can be all kinds of things, including a binomial distribution, twice a binomial distribution, a constant, or nothing resembling a binomial at all.

For example, the following joint distributions all have binomial$(2,\frac12)$ marginal distributions, but the distribution of $X+Y$ varies considerably (representing the four examples listed above): \begin{array}{|c|c|c|} \hline &x=0&x=1&x=2\\ \hline y=0&\frac1{16}&\frac18&\frac1{16}\\ \hline y=1&\frac1{8}&\frac14&\frac1{8}\\ y=2&\hline \frac1{16}&\frac18&\frac1{16}\\ \hline \end{array} \begin{array}{|c|c|c|} \hline \frac14&0&0\\ \hline 0&\frac12&0\\ \hline 0&0&\frac14\\ \hline \end{array} \begin{array}{|c|c|c|} \hline 0&0&\frac14\\ \hline 0&\frac12&0\\ \hline \frac14&0&0\\ \hline \end{array} \begin{array}{|c|c|c|} \hline 0&\frac14&0\\ \hline \frac14&\frac14&0\\ \hline 0&0&\frac14\\ \hline \end{array} The joint distribution that you specified is one of the special cases for which $X$, $Y$ and $X+Y$ are all binomial. This is true not only because $X$ and $Y$ are dependent, but also because they have a specific joint distribution (the multinomial distribution). As an exercise, you can verify that in case $n=2$, the stated joint distribution is the only one for which $X$ has binomial($n,p$), $Y$ has binomial($n,q$) and $X+Y$ has binomial($n, p+q$) distribution.

grand_chat
  • 40,909