There are two random variables $A$ and $B$ with probability distributions given by, $$ \operatorname{Pr}(A=n)=\operatorname{Pr}(B=n)=p(1-p)^{n-1}, \quad 0<p<1, n=1,2, \ldots $$ Define the random variable $C$ to be $\max\left(A,B\right)$. I need to find;
- Average value of $C$
- Average value of $C$ compared to $A$ and $B$ in the limit $p \ll 1$?
Here's what I have so far,
$$\operatorname{Pr}(C=n)=2 \left( p(1-p)^{n-1} \right)\left(\sum_m^{n-1}p(1-p)^{m-1}\right)+\left( p(1-p)^{n-1} \right)\left( p(1-p)^{n-1} \right)$$
The above expression gives us the probability distribution of $C$.
The first term correspond to the probabilities of $A$ (or $B$) being equal to $n$ and $B$ (or $A$) less than $n$. The factor of $2$ corresponds to the choice of $A$ or $B$
The second term is the probability to get both $A$ and $B$ equal to $n$.
Thus average value of $C$ is equal to, $$\sum_n n \operatorname{Pr}(C=n)=\frac{2 p-3}{(p-2) p}$$
Neglecting $p^2$ term in the limit $p\ll 1$, this becomes $\frac{3}{2p}-1$
Average value of $A$ or $B$ is equal to, $$\sum_n n \operatorname{Pr}(A=n)=\frac{1}{p}$$
I calculated all of this using Mathematica.
Is my approach correct?
\llinstead of<<. – K.defaoite Sep 23 '22 at 10:48