# Possible to evaluate definite integral of inverse trigonometric function as function of $Y$?

Suppose $Y = \sqrt{2T}\cos(U)$, $0 \le u \le \pi$, and $0 \le \cos^{-1}(\frac{y}{\sqrt {2t}}) \le \pi )$, so $-1 \le \frac{y}{\sqrt{2t}} \le 1$, with all $\mathbb{R}$. Now I have the iterated integral $$G(y)= \int_{0}^{y}\int_{0}^{\cos^{-1}(\frac{y}{\sqrt {2t}})} \frac{e^{-t}}{\pi } \mathrm{d}u \mathrm{d}t$$

Evaluating the inner integral, I have

$$G(y)=P(Y\le y)=\int_{0}^{y} \frac{e^{-t} \cos^{-1}(\frac{y}{\sqrt{2t}})}{\pi } \mathrm{d}t$$

where $G(y)$ is a CDF of $Y$.

Here I'm stuck. Maybe my integration conditions are erroneous? Any help on how to find this (or the correct) integral as a function of $Y$, would be appreciated.

(This question follows from my previous one: Help solving CDF for transformation of $\ge 2$ random variables or if it's impossible.).

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Once again, you are neglecting the signs since Y might be negative hence the interval (0,y) in the integral should be (-oo,y). – Did Sep 6 '12 at 5:32

Precisely the kind of example where the functional approach explained there works better... Here is how to apply it in the present case.

One may find difficult to juggle with all the inequalities that the computation of the CDF of $Y$ involves. Instead, one can start with any suitable function $\varphi$ and note that, by definition of the distribution of $(T,U)$, $$\mathrm E(\varphi(Y))=\mathrm E(\varphi(\sqrt{2T}\cos(U)))=\int_0^\pi\int_0^\infty \varphi(\sqrt{2t}\cos(u))\frac1\pi\mathrm e^{-t}\mathrm dt\mathrm du.$$ The change of variables $(t,u)\to(y,x)$ with $y=\sqrt{2t}\cos(u)$ and $x=\sqrt{2t}\sin(u)$ yields $2t=x^2+y^2$ and $\mathrm dt\mathrm du=\mathrm dx\mathrm dy$ with $y$ in $(-\infty,+\infty)$ and $x\geqslant0$. Hence, $$\mathrm E(\varphi(Y))=\int_{-\infty}^{+\infty}\varphi(y)\frac1\pi\mathrm e^{-y^2/2}\left(\int_0^\infty\mathrm e^{-x^2/2}\mathrm dx\right)\mathrm dy.$$ The inner integral is independent of $y$ and is equal to $\frac12\sqrt{2\pi}$ hence $$\mathrm E(\varphi(Y))=\int_{-\infty}^{+\infty}\varphi(y)\gamma(y)\mathrm dy,\qquad\text{with}\quad \gamma(y)=\frac1{\sqrt{2\pi}}\mathrm e^{-y^2/2}.$$ This identity holds for every (bounded measurable) function $\varphi$. This proves that the distribution of $Y$ has standard normal density $\gamma$.

Nota: The very same approach starting from $(Y,X)=(\sqrt{2T}\cos(U),\sqrt{2T}\sin(U))$ shows that $X$ and $Y$ are i.i.d. standard normal, a fact which is at the basis of Box-Muller method of simulation of normal random variables.

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If you want the region $0 < u < \pi$, $-1 \le y/\sqrt{2t} \le 1$, you'll need $t \ge y^2/2$, but I don't see why you want $u \le \cos^{-1}(y/\sqrt{2t})$ or $t \le y$ unless there's something you aren't telling us. So if you want to integrate $e^{-t}/\pi$ over this region it should just be

$$\int_{y^2/2}^\infty \int_0^\pi \frac{e^{-t}}{\pi}\ du\ dt$$

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to explain why $t \le y$, I should have included that this integral is to find a CDF function for $Y$. It relates to my previous one: math.stackexchange.com/questions/190809/… – T. Webster Sep 6 '12 at 4:49