We have a generator $X$ that selects numbers from a uniform distribution on $(0,1)$ denoted $\text{Unif}(0,1)$.

We have to show how it can be used with the function $x \mapsto \log\left(\frac{x}{1-x}\right)$ to generate a random number from a distribution with cumulative distribution function $F(x)=\frac{e^x}{1+e^x}$.

The calculation required is to verify that $F^{-1}\left(X\right)$ has cumulative distribution function $F$.

I try it this way:

Let random variable $X \sim \text{Unif}(0,1)$.

$P\left(\log\left(\frac{X}{1-X}\right)\le x\right)=P\left(\frac{X}{1-X}\le e^x\right)=P\left(X\le \frac{e^x}{1+e^x}\right)$.

I think that this is the way but I don't know how to continue and how to justify that:

$P\left(X\le \frac{e^x}{1+e^x}\right)=\frac{e^x}{1+e^x}$.

  • 1
    $\begingroup$ The last step is true because $X$ is uniform on $[0,1]$: $P(X \leq u) = u$ $\endgroup$
    – Adam
    Dec 29 '20 at 23:49

We have $X \sim \text{Unif}(0,1)$, $Y \sim F$, where $F(x) = \frac{e^x}{1+e^x}$.

We want to show that $F^{-1}(X)$ has the same distribution as $Y$. In other words, we must verify that \begin{align*} \mathbb{P}(F^{-1}(X) \leq y) = \mathbb{P}(Y \leq y) = F(y). \end{align*} By straightforward calculations we find that $F^{-1}(x) = \log\big(\frac{x}{1-x}\big)$. Thus \begin{align*} \mathbb{P}(F^{-1}(X) \leq y) &= \mathbb{P}\!\left(\log\left(\frac{X}{X-1}\right) \leq y \right) \\ &= \mathbb{P}\left(X \leq \frac{e^y}{1+e^y}\right) \\ &= \frac{e^y}{1+e^y} = F(y), \end{align*} as desired. We have used that $\mathbb{P}(X \leq x) = x$ for $x\in(0,1)$. This holds since $X \sim \text{Unif}(0,1)$.


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