Convolution of two geometric distributions

[Ad VanderVen]

TL;DR

I am trying to derive the convolution of two geometric distributions but obviously I am making an error.

I'm trying to derive the convolution from two geometric distributions, each of the form:
$$\displaystyle \left( 1-p \right) ^{k-1}p$$ as follows $$\displaystyle \sum _{k=1}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}.$$ with as a result: $$\displaystyle \left( 1-p \right) ^{z-2}{p}^{2}z$$ Now the sum: $$\displaystyle \sum _{z=2}^{\infty } \left( 1-p \right) ^{-2+z}{p}^{2}z.$$ should be equal to $\displaystyle \frac{2}{p}$ which is not the case.

What am I did wrong?

 

[BvU]

$$\displaystyle \sum _{k=1}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}$$ seems strange. You sure it's not $$\displaystyle \sum _{k=1}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k+1}\ ? $$

 

[Ad VanderVen]

The case is treated at: https://math.stackexchange.com/ques...of-two-independent-geometric-random-variables and the summation:$$\displaystyle \sum _{k=1}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}$$ must be$$\displaystyle \sum _{k=1}^{z-1} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}.$$

 

[Stephen Tashi]

I'm trying to derive the convolution from two geometric distributions, each of the form:
$$\displaystyle \left( 1-p \right) ^{k-1}p$$

That indicates that you define the geometric distribution at $k$ to be the probability of success on the $k$-th attempt, as oppose to the probability of success after $k$ failures.

So , as noted on the stackexchange link, if you consider the sum of two geometric random variables, the smallest possible value of $k$ is 2, which results when you succeed on the 1st try from the first distribution and again on the first try from the second distribution. So it isn't clear why you are writing a summation beginning with $k = 1$.

as follows $$\displaystyle \sum _{k=1}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}.$$

 

[Ad VanderVen]

So , as noted on the stackexchange link, if you consider the sum of two geometric random variables, the smallest possible value of $k$ is 2, which results when you succeed on the 1st try from the first distribution and again on the first try from the second distribution. So it isn't clear why you are writing a summation beginning with $k = 1$.

I'm afraid you're absolutely right. Only then will you get $$\displaystyle P \left( X+Y=z \right) \, = \,\sum _{k=2}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}$$

 

[Stephen Tashi]

The case is treated at: https://math.stackexchange.com/ques...of-two-independent-geometric-random-variables and the summation:$$\displaystyle \sum _{k=1}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}$$ must be$$\displaystyle \sum _{k=1}^{z-1} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}.$$

That formula comes from the message on stackexchange that asked the question and conjectured a formula. Look at the formulas given by the messages that answered the question.

$$\displaystyle P \left( X+Y=z \right) \, = \,\sum _{k=2}^{z} \left( 1-p \right) ^{k-1}{p}^{2} \left( 1-p \right) ^{z-k-1}$$

$P(X+Y =2) = p^2$, Does the above formula agree with that when $z = 2$?

$P(X+Y = 3) = p(1-p)p + (1-p)pp = 2(1-p)p^2$