$X$ and $Y$ are geometric, independent random variables with parameters $x$ and $y$ respectively. Use their joint P.M.F. to compute an approximate summation for $P(X<Y)$.

I've setup the problem: $\sum_{i=1}^j \sum_{j=j}^\infty x(1-x)^{i-1} y(y-1)^{j-1}$

But I have no idea how to simplify further from here.

  • 1
    $\begingroup$ X and Y are geometric, independent random variables with parameters x and y respectively. .......... This is most unsavoury notation. If $X$ and $Y$ are random variables, such that $x$ and $y$ are realisations of those random variables, then you cannot use $x$ and $y$ to denote the parameters of the distribution. Try $p$ and $q$ for the parameters ... $\endgroup$ – wolfies May 9 '13 at 20:15
  • $\begingroup$ And the notation $\sum_{j=j}^{\ldots} \ldots$ is a bit inconvenient too... $\endgroup$ – TMM May 9 '13 at 20:32

For a geometric random variable $Y$ with parameter $y$, $$\begin{align} P\{Y > n\} &= \sum_{i=n+1}^\infty P\{Y = i\} = \sum_{i=n+1}^\infty y(1-y)^{i-1}\\ &= y(1-y)^n[1 + (1-y) + (1-y)^2 + \cdots ]\\ &= y(1-y)^n \times \frac{1}{1-(1-y)}\\ &=(1-y)^n. \end{align}$$ Actually, the end result is more easily derived by arguing that the event $\{Y > n\}$ occurs if and only if the first $n$ trials ended in failure, and the probability of this is just $(1-y)^n$. Regardless of which way you computed $P\{Y > n\}$, you can compute $$P\{X = n, Y> n\} = P\{X = n\}P\{Y > n\} = x(1-x)^{n-1}(1-y)^n$$ and hence $$\begin{align} P\{X < Y\} &= \sum_{n=1}^\infty P\{X = n, Y > n\}\\ &= \sum_{n=1}^\infty x(1-x)^{n-1}(1-y)^n\\ &= x(1-y) [1 + (1-x)(1-y) + ((1-x)(1-y))^2 + \cdots\\ &= x(1-y)\times \frac{1}{1 - (1-x)(1-y)}\\ &= \frac{x(1-y)}{x+y-xy}. \end{align}$$ This answer too has a nice interpretation. Let us carry out the $n$-th trials simultaneously. Then, we can ignore all instances in which failures were recorded on both trials -- the next pair of trials must be conducted -- and concentrate on the very first time that success occurred on at least one of the paired trials, possibly on both trials. Conditioned on at least one success, an event of probability $x+y-xy$, (remember $P(A\cup B) = P(A)+P(B)-P(A)P(B)$ for independent events?), the probability that $X < Y$ is just the probability that $X$ occurred and $Y$ did not, which is $x(1-y)$. Thus we have $$P\{X < Y\} = \frac{x(1-y)}{x+y-xy}, ~~ P\{X > Y\} = \frac{(1-x)y}{x+y-xy}, ~~ P\{X = Y\} = \frac{xy}{x+y-xy}$$ without any need for summing series or indeed without even writing down the joint pmf etc.


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