# Exponential distribution $P(Z \geq 5)$

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?

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.

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.

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

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

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

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

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