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I came across the following problem in a book I was reading on continuous probability distributions:-

$Q.$ Let $Y$ be uniformly distributed on $(0,1)$. Find a function $\phi$ such that $\phi(Y )$ has the gamma density $\Gamma(\frac12,\frac12)$.

I know that the probability density represented by $\Gamma(\frac12,\frac12)$ is the following:-

$$\Gamma\left(\frac12,\frac12\right)=\begin{cases}\frac1{\sqrt{2\pi x}}.e^{-\frac x2} &&&& x \ge 0 \\ 0 &&&& x <0\end{cases}$$

I don't have any idea what to do after this. I would also like to have insight on similar questions.

2 Answers 2

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I don't have any idea what to do after this.

So you want $f_{\phi(Y)}(x) = f_Y(\phi^{-1}(x))\cdot\lvert\frac{\mathrm d \phi^{-1}(x)}{\mathrm d x}\rvert$

and know $f_{\phi(Y)}(x) = \sqrt\frac{2}{\pi x}~e^{-2x}~\mathbf 1_{x\in(0;\infty)}$ and $f_Y(\phi^{-1}(x))=\mathbf 1_{\phi^{-1}(x)\in(0;1)}$ (assuming $\phi$ is bijective).

So what you do is substitute, rearrange and integrate.

$$\begin{align} \newcommand{\erf}{\operatorname{erf}} \sqrt\frac{2}{\pi x}~e^{-2x} &= \lvert\frac{\mathrm d \phi^{-1}(x)}{\mathrm d x}\rvert \\[1ex] \phi^{-1}(x) &=\frac{1}{\sqrt{2\pi}}\int_0^x \frac{e^{-s/2}}{\surd s} \operatorname d s &&{x\in(0;\infty)} \\[2ex] y &= \frac{1}{\sqrt{2\pi}}\int_0^{\phi(y)} s^{-1/2}e^{-s/2}\operatorname d s && {y\in(0;1)} \\[1ex] &= \bbox[white,1ex,border:silver 1pt dotted]{\color{white}{\erf(\sqrt{\phi(y)/2~})}} \end{align}$$

and you can take it from there.

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    it seems like the answer is incomplete...could you pls complete the answer? – 2017-02-01
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    You asked for what to do next. This is what you do next. You can complete it from there. – 2017-02-01
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    Ohk...cool down @Graham Kemp...and thanks for the answer – 2017-02-03
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If we assume that $\phi$ is increasing, we can get: $$t=P\left(Y

This is actually a well known result, which is used to generate a random sample from a given distribution. Look up inverse transform sampling.