Consider two independent geometric random variables $$X\sim \text{Geometric}(1/2) \ \ \ \ \text{and} \ \ \ \ Y\sim \text{Geometric}(3/4).$$
My goal is to find
$$\mathbb{P}(X-Y=2).$$
Using the convolution formula
$$\mathbb{P}(X-Y=k)=\sum_{i=1}^{\infty}\mathbb{P}(X=i)\mathbb{P}(Y=i+k),$$
I obtained the sum
$$\mathbb{P}(X-Y=2)=\sum_{i=1}^{\infty}\left ( \frac{1}{2} \right )\left ( \frac{1}{2} \right )^{i-1}\left ( \frac{3}{4} \right )\left ( \frac{1}{4} \right )^{i+1}.$$
This simplifies to the sum
$$\frac{3}{16}\sum_{i=1}^{\infty}\left ( \frac{1}{2} \right )^{i}\left ( \frac{1}{4} \right )^{i}.$$
This is a geometric sum which converges to
$$\frac{\frac{3}{128}}{1-\frac{1}{8}}=\boxed{\frac{3}{112}}.$$
However, I am not too confident about this answer. For example, one of my friends was doing this problem and he used the formula
$$\mathbb{P}(X-Y=k)=\sum_{i=1}^{\infty}\mathbb{P}(X=i+k)\mathbb{P}(Y=i),$$
leading to a different answer.
Can anyone tell me the correct way to approach this? Thanks!
Best Answer
Considered a random variable geometric as the number of trials until the first success: $$Z\sim Geo(p)$$ if $$\mathbb{P}(Z=z)={(1-p)}^zp,\quad z=1,2,3,\dots$$ where $0\le p\le 1$ is the probability of success.
Note that $$ \{X-Y=k\}=\bigcup_{i=k+1}^\infty\{Y=i-k,X=i\}=\bigcup_{j=1}^\infty\{X=j+k,Y=j\}.$$ Then $$ \mathbb{P}(X-Y=k)=\sum_{i=k+1}^\infty \mathbb{P}(Y=i-k,X=i)=\sum_{i=k+1}^\infty \mathbb{P}(Y=i-k)\mathbb{P}(X=i)$$ and $$\mathbb{P}(X-Y=k)=\sum_{j=1}^\infty \mathbb{P}(X=j+k)\mathbb{P}(Y=j).$$