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I am doing some work on finding the distribution of a sum of two independent random variables. In my actual work these two variables are independent, yet have a very different distribution.

I wanted to treat an example case, but I lack the knowledge of double summations. Before starting the example, let me say that I fully know how to obtain the cumulative distribution function in a number of ways, so this question is purely on how to compute double summations of the kind I find in the following example.

Thanks in advance.


Let $X_1 \sim geom(p_1)$, $X_2 \sim geom(p_2)$, both with support $\{1,2,\ldots\}$ and $Y = X_1 + X_2$.

I want to compute the cumulative distribution function of $Y$, using the following approach,

\begin{align} \mathbb{P}(Y \leq k) &= \sum_{j = 1}^{k-1} \sum_{i=1}^{k-j} \mathbb{P}(X_1=i)\mathbb{P}(X_2=j) \\ &= \sum_{j = 1}^{k-1} \sum_{i=1}^{k-j} p_1(1-p_1)^{i-1} p_2(1-p_2)^{j-1}. \end{align}

Here is where I get stuck. How to compute the above double summation? I know, using Mathematica to evaluate the double summation, that the answer should be

\begin{align} \mathbb{P}(Y \leq k) &= \frac{p_1-p_1(1-p_2)^k+p_2((1-p_1)^k-1)}{p_1-p_2}, \end{align}

which is indeed correct when compared to e.g. a generalized negative binomial distribution with correct parameters.

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up vote 2 down vote accepted

I should state that the following steps depend on the formula for a geometric series:

$$\sum_{j=1}^{k} r^{j-1} = \frac{1-r^k}{1-r}$$

The sum in question is

$$p_1 p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} \sum_{i=1}^{k-j} (1-p_1)^{i-1} $$

Evaluate the inner sum first:

$$\sum_{i=1}^{k-j} (1-p_1)^{i-1} = \frac{ 1 - (1-p_1)^{k-j} }{p_1} $$

Now the sum is

$$\begin{align} p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} \left [ 1 - (1-p_1)^{k-j} \right ] &= 1 - (1-p_2)^{k-1} - p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} (1-p_1)^{k-j} \\ \end{align} $$

The sum on the right-hand side may be evaluated as follows:

$$\begin{align}\sum_{j=1}^{k-1} (1-p_2)^{j-1} (1-p_1)^{k-j} &= (1-p_1)^{k-1} \sum_{j=1}^{k-1} \left ( \frac{1-p_2}{1-p_1} \right )^{j-1} \\ &= (1-p_1)^{k-1} \frac{ 1 - \left ( \frac{1-p_2}{1-p_1} \right )^{k-1}}{1 - \frac{1-p_2}{1-p_1}} \\ &= (1-p_1)\frac{(1-p_1)^{k-1} - (1-p_2)^{k-1}}{p_2-p_1} \\ \end{align} $$

Now we can put this all together:

$$\begin{align} 1 - (1-p_2)^{k-1} - p_2 (1-p_1)\frac{(1-p_1)^{k-1} - (1-p_2)^{k-1}}{p_2-p_1} \\ = 1 - \frac{(p_2-p_1)(1-p_2)^{k-1} + p_2 (1-p_1)[(1-p_1)^{k-1} - (1-p_2)^{k-1}]}{p_2-p_1} \\ \end{align}$$

which, after some cancellation and consolidation, produces the following result for the sum:

$$1 - \frac{p_1 (1-p_2)^k - p_2 (1-p_1)^k}{p_1-p_2}$$

and is equal to the result stated above.

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Thank you for you elaborate response, much appreciated. –  user60307 Jan 31 '13 at 13:08
    
You're welcome! To be honest, I like the form of the result I posted better than what you posted from Mathematica. You can see the symmetry in the behavior of the sum with respect to $p_1$ and $p_2$. Anyway, that was a lot of fun; feel free to post any more like this. –  Ron Gordon Jan 31 '13 at 13:25
    
I agree, it clearly shows that when $k$ increases, the term on the right goes to zero and the asymptote is 1. Thank you for your comments. –  user60307 Jan 31 '13 at 13:30
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