Exponential distribution $P(Z \geq 5)$

Question

Exponential distribution

Let $Z ∼ Exponential(4)$. Compute each of the following

(a) $P(Z \geq 5)$

$$P(Z \geq 5) = \int_{5}^{\infty} 4e^{-4x}dx$$

Let $u = -4x$, then $du = -4dx \leftrightarrow -\frac{1}{4}du = dx$

$$-\int_{-\infty}^{-20} e^{u} du = -e^{u}|_{-\infty}^{-20} = -(e^{-20} - \lim_{u\to-\infty}e^{A}) = -e^{-20} + 0 = -e^{-20}$$

Answer is $e^{-20}$. Where did I go wrong or is the solution wrong?

Answer 1

It should be pointed out that the notation $$Z \sim \operatorname{Exponential}(4)$$ is imprecise, because it does not tell us whether the $4$ is a rate parameter (as you have implied in your computation), or a scale parameter. That is to say, does the above mean $$\operatorname{E}[Z] = 1/4, \quad f_Z(z) = 4e^{-4z}, \quad z \ge 0,$$ or does it mean $$\operatorname{E}[Z] = 4, \quad f_Z(z) = e^{-z/4}/4, \quad z \ge 0?$$ The probability $\Pr[Z \ge 5]$ will be quite different depending on the parametrization.

For the sake of completeness, we should observe that $$\Pr[Z \ge z] = S_Z(z) = 1 - F_Z(z) = 1 - (1 - e^{-\lambda z}) = e^{-\lambda z},$$ if parametrized by rate, and $$\Pr[Z \ge z] = e^{-z/\theta}$$ if parametrized by scale.

This ambiguity also carries over into other members of the exponential family; e.g., the gamma distribution. Both parametrizations are used in the literature and there is no universally accepted preference of rate versus scale.

Answer 2

$P(Z \ge 5)=1-P(Z \lt 5)$

$=1-(1-e^{-20})=e^{-20}$

Answer 3

First of all, the solution is obviously wrong. You've got a negative probability.

The problem is in the $$-\int\limits_{-\infty}^{-20}$$ step.

Your original range was from $5$ till $\infty$. You substituted a new variable, $u = -4x$, hence its range is from $-20$ to $-\infty$, i.e.

$$P(Z \geq 5) = \int\limits_5^{\infty}4e^{-4x}\mathrm{d}x = -\int\limits_{-20}^{-\infty}e^u\mathrm{d}u = \int\limits_{-\infty}^{-20}e^u\mathrm{d}u = e^{-20}$$

Answer 4

Just this: $$\int_5^\infty 4 e^{-4x}\mathsf d x= \lim_{x\to\infty}(-e^{-4x})-(-e^{-4\cdot 5})$$

When you apply the substitution

$$\begin{split}\int_{-20}^{-\infty} 4e^{u}\dfrac{\mathsf d u}{-4} &= -\int_{-20}^{-\infty}e^u\mathsf d u \\ &=\int_{-\infty}^{-20}e^u\mathsf d u \\ &=e^{-20}-\lim_{u\to-\infty}e^u\end{split}$$