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Consider $X∼unif [0,1]$. Find a function $g: \mathbb{R} \longrightarrow \mathbb{R}$, such that g(X) has pdf $f(t) = \begin{cases} {t+1}, & \text{$-1 \leq t\leq 0$} \\ {1-t}, & \text{$0<t\leq 1$}\end{cases}$.

Can you help me, please? I do not know what I have to do.

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Hint/Guide

Compute the cdf $F$ corresponding to $f$ (i.e. $F(x)=\int_{-\infty}^x f(t)\, dt)$ and use the fact that $F^{-1}(X)$ has the same distribution as $F$. So we can take $g=F^{-1}$.

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  • $\begingroup$ Thanks for your answer. So I have $ F(x) = \begin{cases} 0 & x <-1\\ \frac12 + x + \frac{x^2}{2} & x \in[-1,0]\\ \frac12 + x - \frac{x^2}{2} & x \in(0,1]\\ 0 & x > 1\end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right? $\endgroup$ – tommy_m Jan 3 at 18:36

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