problem 43

5.3.3 problem 43

a. The first success occurs at times $k Δ t$ , where $k \geq 0$ is the number of failures before the first success. Therefore, $T = G Δ t$ .

b. Since the trials are independent and with the same probability of success $λ Δ t$ , the number of failures before the first success is $G \sim Geom (λ Δ t)$ by the story of the geometric. The PMF of $G$ is given by

P (G = k) = {(1 - λ Δ t)}^{k} λ Δ t, for k = 0, 1, \dots

It is convenient to calculate $P (G > k)$

P (G > k) = \infty \sum i = k + 1 P (G = i) = λ Δ t \infty \sum i = k + 1 {(1 - λ Δ t)}^{i} = {(1 - λ Δ t)}^{k + 1}

The CDF of $T$ is calculated as

\begin{matrix} P (T \leq t) & = 1 - P (T > t) = 1 - P (G > t ∕ Δ t) = 1 - {(1 - λ Δ t)}^{\frac{t}{Δ t} + 1} \end{matrix}

c. Taking the limit of the CDF of $T$ for $Δ t \to 0$

lim Δ t \to 0 P (T \leq t) = lim Δ t \to 0 [1 - {(1 - λ Δ t)}^{\frac{t}{Δ t} + 1}] = 1 - lim Δ t \to 0 {(1 - λ Δ t)}^{\frac{t}{Δ t} + 1}

If we do the substitution $n = t ∕ Δ t$ , we obtain $n \to \infty$ for $Δ t \to 0$

lim Δ t \to 0 P (T \leq t) = 1 - lim n \to \infty {(1 - \frac{λt}{n})}^{n + 1} = 1 - lim n \to \infty {(1 - \frac{λt}{n})}^{n} \times lim n \to \infty (1 - \frac{λt}{n})

The first limit is the compount interest limit and is equal to $e^{- λt}$ . The second limit is equal to 1. Thus,

lim Δ t \to 0 P (T \leq t) = 1 - e^{- λt}

which is the CDF of the $Expo (λ)$ distribution. Therefore, $T$ converges to $Expo (λ)$ as trials are performed faster and with smaller success probabilities.

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