Let $X$ and $y$ be independent, identically distributed random variables with probabilities ...

Question

Let $X$ and $y$ be independent, identically distributed random variables with probabilities

$$P(X=n)=\left\{\begin{matrix} pq^{n-1}, n\geq 1\\ 0, n<1 \end{matrix}\right.$$ $$P(Y=m)=\left\{\begin{matrix} pq^{m-1}, m\geq 1\\ 0, m<1 \end{matrix}\right.$$
where $0<p<1$ and $p+q=1$. Evaluate the probabilities of the sum of $X$ and $Y$ $$P(Z=r)=P(X+Y=r).$$

My solution so far:

Note that $P(X=n)=f_{X}(n)$ and $P(X=m)=f_{Y}(m)$.

$$P(Z=r) = P(Z=X+Y) = f_{Z}(z).$$ $$f_{Z}(z) = \sum_{{\left \{ (n,m)|n+m=r \right \}}}P(X=n,Y=m)$$ $$=\sum_{n}P(X=n,Y=z-n)=\sum_{n}f_{X}(n)f_{Y}(z-n)$$ $$=\sum_{n}pq^{n-1}pq^{z-n-1}= p^{2}q^{z-2}\sum_{n}q^{0}= p^{2}q^{z-2}\sum_{n=1}^{r-1}1 = p^{2}q^{z-2}(r-1).$$

EDIT - I believe I got it now.

It seems that the sum diverges to infinity, which is not correct. Is there anything wrong here?

Thanks for answering!!

Answer 1

• You did not consider the cases when $z-n < 1$, of which its pmf gets evaluated to $0$.

• You seems to use two notations, $z$ and $r$ to denote the same thing.

\begin{align} \sum_{n=1}^\infty f_X(n) f_Y(r-n)&=\sum_{n=1}^{r-1} f_X(n) f_Y(r-n) \end{align}

which is a finite sum.

Answer 2

Your sum is bounded because the number of terms is finite. You can only sum from $n=1$ to $n=r-1$, otherwise $Y$ would drop out of the supported range.

Answer 3

... Greetings, robot friend!

The reason why you are encountering a DOES NOT COMPUTE is that some programmer has mistakenly forgotten to bound one of your loops. The sum cannot go larger than $z-1$ (remember that $n$ is the value of $X$ which can't be larger than $z-1$ since $Z=X+Y.$) So instead you need to take $\sum_{n=1}^{z-1}.$