Relation between the hazard rate and the expected value of an exponential random variable In a certain question about expectation, I am given that a certain RV, "follows an exponential distribution with a hazard rate = 1". I am told nothing else about this distribution. I am also given later that the expected value for this distribution = 1. Why is this?
 A: In survival analysis, let there be a random variable $X$, we define it's hazard function as
\begin{align}
\lambda_{t} = -\frac{S'(x)}{S(x)}
\end{align}
where $S'(x)$ denote the first derivative of $S(x)$ and $S(x)$ is the survival function given by
\begin{align}
S(x)=1-F(x)
\end{align}
where $F(x)$ is the cumulative distribution function of $X$.
If $X$ is an exponential random variable with parameter $\mu$ then it's survival function takes the following form:
\begin{align}
F(x) &= 1-e^{-\mu x}\\
S(x) &= e^{-\mu x}
\end{align}
where $\mu>0$. Let's calculate the hazard now:
\begin{align}
S'(x)= -\mu e^{-\mu x}\\
\lambda(x) = \frac{\mu e^{-\mu x}}{e^{-\mu x}}=\mu
\end{align}
You can see that $S(x)=1=\mu$. Now find the formula for the expected value of $X$:
\begin{align}
\mathbb{E}(x) &= \int_0^\infty x\mu\exp(-\mu x)\,dx\\
              &= \left(\left.-\frac{\mu}{\mu}e^{-\mu x}x - \frac{\mu}{\mu^2}e^{-\mu x} \right)\right|^\infty_0\\
&= \frac{\mu}{\mu^2} = \frac{1}{\mu}
\end{align}
Given that $\mathbb{E}(X)= \frac{1}{\mu}$, we know that $\mathbb{E}(X)=1$.
A: For the exponential distribution $\operatorname{Exp}(\lambda)$, its hazard rate coincides with the parameter $\lambda$.
In general, if $S$ is a survival function (for a continuous time of event $T$)
$$1-F_T(t)=:S_T(t)=\exp\Big(\int^t_0 h_T(s)\,ds\Big)$$
where $h_T$ is the hazard function.
For an exponential random variable $T$ with parameter $\lambda$
$$S_T(t)=e^{-\lambda t}$$
where one concludes that $\int^t_0 h_T(s)\,ds =\lambda t$ and so, $h_T(t)\equiv\lambda$ for all $t>0$.
