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There is a discrete process that yields one of 3 outputs at each trial: $A$ with probability $p_{A}$, $B$ with $p_{B}$ and $STOP$ with probability $1 - p_{A} - p_{B}$ (the process terminates after yielding $STOP$). Let $X$ and $Y$ be a number of $A$'s and $B$'s we get till the process terminates. Then

a) compute $E(X), Var(X)$ ... It smells like quasi-geometric distribution but I can't figure out how to attack it as it is potentially infinite ...

b) find the smallest $a > 0$ such that $P(X > n) = O(a^{n})$

c) are $X$ and $Y$ independent?

For b), how to get $P(X > n)$ or $P(X <= n)$?

For c) I think the easiest way is to check if $P(X=k, Y = l) = P(X=k)P(Y=l)$ for all $k, l$. LHS is straightforward as I know when the process stops so it is $\binom{k+l}{k}p_{A}^{k}p_{B}^{l} = (p_{A} + p_{B})^{k+l}$ but how to get $P(X=k)$? (so pretty the same issue as for b))

Thank you in advance

1 Answers 1

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Yes, it is very similar to the Geometric Distribution.

We find the expected count of $A$ events before STOP, by partitioning on the result of the first roll :

$$\begin{align}\mathsf E(X) ~&=~ p_A~\mathsf E(1+X) + p_B~\mathsf E(X)+ (1-p_A-p_B)~0 \\[1.5ex] (1-p_A-p_B)~\mathsf E(X) ~&=~ p_A \\[1ex] \mathsf E(X) ~&=~\dfrac{p_A}{1-p_A-p_B}\end{align}$$

Then also: $\mathsf E(Y)= \dfrac{p_B}{1-p_A-p_B}$

Now, can you tackle the variance?

$$\require{cancel}\mathsf E(X^2) = p_A~\mathsf E\big((1+X)^2\big)+p_B~\mathsf E(X^2)~\cancel{+(1-p_A-p_B)~0}$$

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    Many thanks Graham,couldn't be simpler ;) Is there any theorem for this kind of "partitioning" or in general tackling the problems where the number of experiments is random itself? Or this is just straightforward approach? – 2017-01-25
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    It's just an application of the Law of Total Expectation. (The Tower Property.) It is mostly useful in situations where the "memoryless" property comes into play; where $\mathsf E(X\mid X\geqslant n)=n+\mathsf E(X)$. – 2017-01-26