Find $\phi$ such that $\phi(X)\sim Γ(\frac{1}{2},\frac{1}{2})$, for $X\sim U(0,1)$ So here is my attempt: Trying to use the following theorem to find such $\phi$: If $X$ has density function $f_{X}$ and $Y=\phi(X)$ and $\phi$ is strictly monotonic and differentiable function, then $f_{Y}(y)=f_{X}(\phi^{-1}(y))|\frac{dx}{dy}|$ is a density function of $Y$. Let's first look for an increasing $\phi$. After some calculations, we conclude that: $\phi^{-1}(y)=\sqrt{\frac{1}{2\pi}}\int \frac{1}{\sqrt{y}}e^{\frac{-y}{2}}dy$ from which, after substitution of $u=\sqrt{y}$, we obtain $\phi^{-1}(y)=2Φ(\sqrt{y})$, where $Φ$ is the standard normal distribution. We finally obtain $\phi(x)=[Φ^{-1}(\frac{x}{2})]^{2}$ for $x\in{(0,1)}$. But the solution guide indicates that $\phi(x)=[Φ^{-1}(x)]^{2}$ for $x\in{(-1,1)}$. What am I doing wrong ? Thank you in advance.
 A: I'll pick up where I think things may have gone astray.
With $\phi^{-1}$ assumed to be increasing, you have
$$ \frac{e^{-y/2}}{\sqrt{2\pi} \sqrt{y}} = \frac{d}{dy} \phi^{-1}(y) $$
These are both functions of $y$ and are equal for $y > 0$. We'll see momentarily why it's important to note this domain restriction.
The LHS does look a lot like it can be transformed into the pdf of a standard normal random variable.  But to get that $\phi^{-1}$ out of the differentiation operator, it's not a matter of simply finding the antiderivative of both sides.  Note that $\Phi(z) \equiv P(Z\leq z)$ -- i.e.,
$$\Phi(z) = \int_{-\infty}^z f(t) dt, $$
where $f(t)$ is the pdf of a standard normal random variable.  Therefore, if we want to use the normal cdf, we have to take definite integrals of both sides.  But whereas $\Phi(z)$ is defined for all of $\mathbb{R}$, we're only looking at positive numbers.  And so --
\begin{align}
\int_0^t \frac{e^{-y/2}}{\sqrt{2\pi} \sqrt{y}} dy &= \int_0^t \frac{d}{dy} \phi^{-1}(y) dy \\
\iff \; 2 \int_0^\sqrt{t} \frac{e^{-u^2/2}}{\sqrt{2\pi}} du &= \left[ \phi^{-1}(y) \right]_0^t  \\
\iff \; 2 \left[ \Phi(\sqrt{t}) - 1/2 \right] &= \phi^{-1}(t) - \phi^{-1}(0)
\end{align}
If $\phi^{-1}$ is increasing, we can figure out the exact value of $\phi^{-1}(0)$ by looking at the domain and range required for our transformation (or its inverse).  Once we do that, we'll have our answer :).
Note that, even though everything in the end is written in terms of $t$, those are actually values from our gamma distribution -- i.e., they're really $y$s.  Likewise, $\phi^{-1}(t)$ is an $x$-value.

Here's what the inverse transformation looks like -- a uniform distribution, I think, would be a bit more recognizable than a $\Gamma(1/2,1/2)$ -- as simulated in R:
> y <- rgamma(10000, .5, .5)
> x <- 2 * pnorm(sqrt(y)) - 1
> hist(x, freq=FALSE)


