TITLE: Cumulative distribution function of sum of binomial random variables
QUESTION [9 upvotes]: I was wondering how to get the cumulative distribution function of a sum of two random binomial variables.
X + Y, where X has n=15 trials and Y has m=15 trials and probability=0.2 for both 
P(15<= X+Y <= 20) how would I express this as a cumulative distribution function?
Can I just add their probability mass functions Px(X) and Py(Y) and then get the CDF from there or do I need to go from joint distribution formulas? 
Any help would be greatly appreciated!
Thanks.

REPLY [6 votes]: The binomially distributed random variable $X$ records the number of successes if we repeat an experiment independently $m$ times, with probability of success each time equal to $p$. The random variable $Y$ records the the number of successes in $n$ independent trials, and we are told that $X$ and $Y$ are independent.
Define Bernoulli random variables $W_1, W_2, \dots, W_m, W_{m+1}, W_{m+2}, \dots, W_{m+n}$ as follows. Perform an experiment independently $m+n$ times, where the probability of success each time is $p$. 
Let 
$$W=\sum_{i=1}^{m+n} W_i.$$
Then $W$ has binomial distribution with parameters $m+n$, $p$.
Note that $S=\sum_{i=1}^m W_i$ has binomial distribution with parameters $m$, $p$, and that $T=\sum_{i=m+1}^{m+n} W_i$ has binomial distribution with parameters $n$, $p$. So $S$ has the same distribution as $X$, and $T$ has the same distribution as $Y$. 
Note also that $S$ and $T$ are independent. Since the distribution of $X+Y$ is completely determined by the distributions of $X$ and $Y$, we conclude that $X+Y$ has the same distribution as $S+T$. But $S+T=W$, so $X+Y$ is binomially distributed with parameters $m+n$, $p$.   
Remark: Informally, $X+Y$ records the total number of successes in $m+n$ independent trials, where the probability of success on any trial is $p$. So it is "obvious" that $X+Y$ has binomial distribution, and there is nothing much to the argument above. However, showing that if $X$ and $Y$ are independent, then the Bernoulli components of $X$ and $Y$ are independent looks as if it may require some work.  That was the reason for the workaround that used the fact that the distribution of $X+Y$ is completely determined by the individual distributions.