Question

Problem 2. This problem investigates the similarity between the geometric and ex- ponential random variables observed last week. Let Y be a geometric random variable with parameter p, so that Y represents the trial number of the first success in a sequence of independent Bernoulli trials. Suppose the trials occur at times , 2, . . . , and that δ and p are both very small. Let λ /δ. At time t, about t/6 trials have taken place (a) Compute P(Y > m), which represents the probability that no success has been observed by time t to e-M as p, δ 0 with λ p/S fixed. variable with parameter λ. (b) Show that the probability that no success has been observed by time t converges (c) Conclude that the time of the first success is approximately an exponential random

Answer 1

(a)

At time t = m$\delta$, total number of trials is t/$\delta$ = m$\delta$ / $\delta$ = m

By CDF of geometric distribution, probability of no success till m trials is,

$P(Y > m ) = (1 - p)^m$

(b)

Let T be the time of first success. Then the probability that no success has been observed by time t is,

$P(T > t) = P(Y > m)$

As, $\delta \rightarrow 0$ and $p \rightarrow 0$

$P(T > t) = P(Y > m) = \lim_{\delta \rightarrow 0 ~p \rightarrow 0} (1 - p)^m = \lim_{\delta \rightarrow 0 } (1 - \delta \lambda)^m$ {$\lambda = p /\delta$}

$= \lim_{\delta \rightarrow 0 } (1 - \delta \lambda)^{t/\delta}$ {t = m$\delta$}

$= \lim_{\delta \rightarrow 0 } (1 + (-\delta \lambda))^{(-t\lambda)/((-\delta \lambda))}$

$=e^{-t \lambda}$ {  $\lim_{x\rightarrow 0} (1 + x)^{k/x} = e^k$ }

(c)

From part (b),

$P(T > t)=e^{-t \lambda}$

$\Rightarrow P(T \le t)=1 - e^{-t \lambda}$

$\Rightarrow F(t)=1 - e^{-t \lambda}$ where F(t) is the CDF of T.

PDF of t is,

$f(t)= F'(t)=\lambda e^{-t \lambda}$

which is the PDF of exponential distribution with the parameter $\lambda$.

Thus, the time of first success T is approximately an exponential distribution variable with parameter $\lambda$​​​​​​​.