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Suppose independent random variables $U, T$. Let $U$ have continuous Uniform distribution over $(0, 2\pi)$. Let $T$ have Exponential distribution with $\lambda = 1$. Let random variable $Y$ be a function $Y=r(U, T) = \sqrt{2T}\cos(U)$. I am trying to find the CDF of $Y$ (without assuming a direct one-to-one transformation to find the PDF by taking the Jacobian).

$$P(Y \le y) = P(r(U,T) \le y) =\int_{0}^{y}\int_{0}^{\frac{y^2}{2\cdot cos^2(u) }}f_U(u)f_T(t) \mathrm{d}t \mathrm{d}u $$ The random variables $X, Y$ are independent, so their joint PDF is the product of their marginal PDFs; $\frac{e^{-t}}{2 \pi }$. $$ = \int_{0}^{y}\int_{0}^{\frac{y^2}{2\cdot cos^2(u) }}\frac{e^{-t}}{2 \pi } \mathrm{d}t \mathrm{d}u$$

Computing the inner integral, I have $$\int_{0}^{y} -\frac{e^{-\frac{1}{2} \text{Sec}[u]^2 y_1^2}-1}{2 \pi }\mathrm{d}u$$

and I'm stuck here. This integral seems impossible to evaluate to a function of y. Is this the case, or perhaps I have made a mistake in my limits of integration?

T. Webster
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1 Answers1

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The domain of integration of the first double integral written in the post does not correspond to the $(u,t)$-domain $r(u,t)\leqslant y$, since one replaces erroneously the condition $\sqrt{2t}\cos u\leqslant y$ by the pair of conditions $0\leqslant 2t\cos^2u\leqslant y^2$ and $0\leqslant u\leqslant y$ (the first double inequality neglects the signs, for example $r(\pi,1)\leqslant0$ while $2\cdot1\cdot\cos^2(\pi)\gt1$, and the second double inequality is rather bizarre).

All in all, these troubles seem like a good motivation to use the usual method.

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  • I will probably mark yours as the answer (by elimination), although it still hasn't led me to the solution, but has helped me correct my erroneous integration conditions. – T. Webster Sep 06 '12 at 02:02
  • my follow-up question(s) has been posted at http://math.stackexchange.com/questions/191714/possible-to-evaluate-definite-integral-of-inverse-trigonometric-function-as-func – T. Webster Sep 06 '12 at 02:21