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Supposed we are given a random variable $X$, whose probability density function is: $$ f_X(x)=\begin{cases} λ \exp⁡(-λx), & x≥0 \\ 0, & x<0. \end{cases} $$ We wish to find an invertible function $g$, such that $Y=g(X)$ is uniformly distributed on the unit interval $[0,1]$. Find $g(x)$. My answer: Since $Y$ is uniformly distributed on $[0,1)$, $fy =1$. since $fy(y)=fx(x)/|d/dx(g(x))|;$ $d/dx(g(x))=fx(x)$. Therefore, $g(x)= \exp(-λx).$ I am not sure whether this is correct. If it is not, could anyone help me out? Thank you very much.

Michael Hardy
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Deepsea234
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  • please use Mathjax to make your expressions readable. See Michael Hardy's edit to learn how. – spaceisdarkgreen Nov 03 '17 at 01:55
  • There is a reason why, in the notation $f_X(x),$ one distinguishes between capital $X$ and lower-case $x$ and one is careful about which is which. Without that, you could not understand $\Pr(X\le x),$ and a variety of other things. And you get sloppy about that. – Michael Hardy Nov 03 '17 at 01:56
  • Sorry about that. I am a new user and I will learn how to use Mathjax to ask questions clearly. Thank you very much. – Deepsea234 Nov 03 '17 at 02:38

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Let $g:[0,\infty)\to[0,1]$ be the map $u\mapsto 1-e^{-\lambda x}$. Then for $0\leqslant u\leqslant 1$ we have \begin{align} \mathbb P(g(X)\leqslant u) &= \mathbb P\left(1-e^{-\lambda X}\leqslant u\right)\\ &= \mathbb P\left( e^{-\lambda X}\geqslant 1-u \right)\\ &= \mathbb P(-\lambda X\geqslant\log(1-u))\\ &= \mathbb P\left(X\leqslant \frac1\lambda \log(1-u) \right)\\ &= 1-e^{-\lambda\left(\frac1\lambda \log(1-u)\right)}\\ &= 1-(1-u)\\ &= u, \end{align} so that $g(X)$ has $U[0,1]$ distribution. The inverse of $g$ is the map $u\mapsto -\frac1\lambda\log(1-u)$ for $0\leqslant u\leqslant1$.

Math1000
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