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Let $X$ be a random variable taking positive values with the density function $f$.
Let $F(t) = P(X\leq t)$ be distribution function of $X$ and $$ h(t) = \int_0^{t} \frac{f(s)}{1-F(s)}\,ds$$ the so-called cumulative hazard function of $f$.

How could I show that $h(X)$ is exponentially distributed with parameter $\lambda = 1$?

I would be grateful for any ideas or suggestions. Thanks.

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A standard result says that if $Y$ is a random variable whose c.d.f. is continuous, then $F_Y(Y)$, the cumulative probability distribution of $Y$ evaluate at the random variable $Y$, is uniformly distributed on $(0,1)$, so that $\Pr(F_Y(Y) \le y)=y)$ if $0\le y\le1$. $$ \Pr(F_Y(Y)\le y) = \Pr(Y\le F_Y^{-1}(y)) = F_Y\Big(F_Y^{-1}(y)\Big) = y. $$ Some difficulties arise when $F_Y$ is not one-to-one, but they can be overcome and have probably been written up in some other question posted here. I'll see if I can find it.

Now \begin{align} \Pr(h(X)\ge x) & = \Pr\left( \int_0^X \frac{f(s)}{1-F(s)}\,ds \ge x \right) = \Pr\left( \int_1^{1-f(X)} \frac{-du}{u} \ge x \right) \\[8pt] & = \Pr\left( -\log(1-F(X)) \ge x \right) = \Pr(F(X) \ge 1-e^x) \\[8pt] & = e^x. \end{align}

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$$\int_{0}^{t}\frac{f(s)}{1-F(s)}~ds=\int_{0}^{t}h(s)~ds$$ $$\Rightarrow [-\log (1-F(s))]_{0}^{t}=\int_{0}^{t}h(s)~ds$$ $$\Rightarrow -\log[1- F(t)]+\log [1-F(0)]=\int_{0}^{t}h(s)~ds$$ $$\Rightarrow -\log[1- F(t)]=1-\exp[-\int_{0}^{t}h(s)~ds]$$ In particular, if $h(s)=\theta \forall s >0$, then $$F(t)=1-\exp[-\int_{0}^{t}\theta~ds]=1-\exp[-\theta t],t>0$$ Then the PDF of T is $$f(t)=\theta e^{-\theta t},t>0$$ which is the PDF of Exponential Distribution with mean $\frac{1}{\theta}$. Put $\theta=1$ in your problem.

Note: $$\lim_{dt \to 0}\frac{P[t<T<t+dt|T>t]}{dt}=\frac{F'(t)}{1-F(t)}=\frac{f(t)}{1-F(t)}=H(t)$$ is known as instantaneous failure rate or hazard rate at time t.

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