Cumulative distribution function of sum of binomial random variables 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.
 A: Assuming that $X$ and $Y$ are independent, you can use the following, standard, result:
Let $X_1$ and $X_2$ be discrete, independent  random variables with 
${\rm dist }(X_1)={\rm B}(n,p)$ and ${\rm dist }(X_2)={\rm B}(m,p)$, where $B(n,p)$ denotes the Binomial distribution with $n$ trials and success factor $p$.
By independence, it would seem that the total number of successes in $n$ trials of $X_1$ and $m$ trials of $X_2$ should be a binomial variable with parameters $n+m$ and $p$.  We now show that this is, indeed, the case.
Let $Y=X_1+X_2$.
We will find the probability mass function of $Y$.  Since $Y$ is the total number of successes in $n$ trials of $X_1$ and $m$ trials of $X_2$, the random variable $Y$ takes the values $0$, $1$, $\ldots\,$, $n+m$.  Using the Convolution Theorem, for $0\le k\le n+m$, we have:
$$
  \eqalign{
      p_Y(k)&=\sum_{i=0}^kP[X_1=i,X_2=k-i]\cr
      &=\sum_{i=0}^kP[X_1=i]\cdot P[ X_2=k-i]\cr
      &=\sum_{i=0}^k{n\choose i}(1-p)^{n-i}p^i\cdot{m\choose k-i}(1-p)^{m-(k-i)}p^{k-i}\cr
   %   &=\sum_{i=1}^k{n\choose i}{m\choose k-i}(1-p)^{m+n}p^{k}\cr
      &=(1-p)^{m+n-k }p^{k}\sum_{i=0}^k{n\choose i}{m\choose k-i}\cr
      &={m+n\choose k}(1-p)^{m+n-k}p^{k }.
}
$$
Thus, ${\rm dist }(X_1+X_2)={\rm B}(n+m,p)$. 
In the above, we used the following:
Lemma 
For any positive integers  $n$, $m$, and $k\le n+m$:
$$
  \sum_{i=0}^{k} {n\choose i}{m\choose k-i}  = {n+m\choose k}.
$$
Proof:  Apply the Binomial Theorem to the equality
$$
  (1+x)^n(1+x)^m=(1+x)^{n+m}
$$
to obtain
$$\tag{1}
  \sum_{i=0}^n{n\choose i}x^{n-i}\cdot\sum_{j=0}^m{m\choose j}x^{m-j}=\sum_{k=0}^{n+m}{n+m\choose k}x^{n+m-k}. 
$$
But
$$
  \eqalign{
      \sum_{i=0}^n{n\choose i}x^{n-i}\cdot\sum_{j=0}^m{m\choose j}x^{m-j}  
      &=\sum_{i=0}^n{n\choose i}\cdot\Bigl[\sum_{j=0}^m{m\choose j}x^{m-j}\bigr]x^{n-i}\cr
      &=\sum_{i=0}^n\Bigl[\sum_{j=0}^m{n\choose i}{m\choose j}x^{n+m-(i+j)}\Bigr].\cr 
    }
$$
Now, terms of the form $x^{n+m-k}$  on the right hand side of the above equality are obtained only when $0\le i\le k$ and  $j=k-i$. 
Thus, the $x^{n+m-k}$-th term of the left hand side of equation $(1)$ is:
$$
  \sum_{i=0}^k{n\choose i}{m\choose k-i}x^{m+n-k}.     
$$
Since the   $x^{n+m-k}$-th term of the right hand side of equation $(1)$ is
$$
  {n+m\choose k}x^{n+m-k},
$$
we have
$$
  \sum_{i=0}^k{n\choose i}{m\choose k-i}={n+m\choose k},   
$$
as desired. 

Convolution Theorem:
The probability mass function of the sum of two independent discrete variables is the convolution of their  probability mass functions:
Let $X_1$ and $X_2$ be independent, discrete random variables that take integer values with respective probability mass functions $p_{X_1}$ and $p_{X_2}$. Let $Y=X_1+X_2$.
Then for each admissable $k$: 
$$
   p_Y(k)=\sum_{i\le\, k}p_{X_1}(i)p_{X_2}(k-i).
$$
The sum appearing on the right hand side of the above equality is called the convolution of $p_{X_1}$ and $p_{X_2}$.
Proof: Exercise.
A: 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. 
