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Let $X,Y,Z$ be independent random variables with uniform distribution on $[0,1]$. How should I calculate $$ P(Z<\sin X\cos Y)? $$ What if $X,Y,Z$ are of normal distribution $N(\mu,\sigma^2)$? What I've attempted so far is that $$ P(Z<\sin X\cos Y)=P(Z-\sin X\cos Y<0) $$ It seems that we need to find the distribution function the the random variable $$ W=Z-\sin X\cos Y. $$ If we have the distribution of $V=-\sin X\cos Y$, then we can use the convolution of the distribution measure. How can I go on?

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Assume the underlying probability space (we don't need it eventually though) is $(\Omega,\mathcal{A},P)$. Then $$ V=(X,Y,Z):\Omega\to(\Bbb{R}^3,\mathcal{B}_{{\Bbb R}^3}) $$ is a random variable (${\mathcal B}_{\Bbb R^3}$ is the Borel $\sigma$-algebra on ${\Bbb R^3}$). Note that $$ P^V(A)=P(V^{-1}(A)), \quad A\in{\mathcal B}_{\Bbb R^3} $$ is a probability measure on $({\Bbb R}^3,{\mathcal B}_{\Bbb R^3})$ and we have $$ P^V(A)=\int_{A}dP^V=\int_{V^{-1}(A)}dP=P(V^{-1}(A)). $$ This is the reason why we don't need to specify the sample space when we want to calculate $P^V(A)$. One the other hand, when $V$ has a probability density with respect to the Lebesgue measure $\mu$ on $({\Bbb R}^3,{\mathcal B}_{\Bbb R^3})$, we have $$ P^V(A)=\int_Af(x,y,z)d\mu,\quad A\in{\mathcal B}_{\Bbb R^3}. $$ In your question, $A=\{(x,y,z)\in{\Bbb R^3}:z<\sin x\cos y\}$. Since $X,Y,Z$ are independent, the density function of random vector $V=(X,Y,Z)$ $$ f(x,y,z)=f_X(x)f_Y(y)f_Z(z) $$ where $f_X,f_Y,f_Z$ are the density of $X,Y,Z$ respectively. Now you can go on with doing Riemann integrations.

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Consider $X$, $Y$, $Z$ iid random variables with uniform distribution on $[0,1]$. Then: $$ \mathbb{P}\left(Z< \sin(X) \cos(Y) \right) = \mathbb{E}\left(\mathbb{P}\left(Z< \sin(X) \cos(Y) | X,Y \right)\right) = \mathbb{E}\left( F_Z\left(\sin(X)\cos(Y)\right) \ \right) $$ Since $F_Z(z) = \min(1,\max(0,z))$, and since $0 \leqslant \sin(X) \leqslant \sin(1) < 1$ and $0<\cos(1) \leqslant \cos(Y) \leqslant 1$ with probability 1, we have $$ \mathbb{E}\left( F_Z\left(\sin(X)\cos(Y)\right) \ \right) = \mathbb{E}\left( \sin(X)\cos(Y) \right) = \underbrace{\mathbb{E}\left(\sin(X)\right)}_{} \, \mathbb{E}\left(\cos(Y)\right) $$ Remaining expectations are easily computed directly.

If $X$, $Y$ and $Z$ are iid standard normal random variables, then $F_Z(z) = \Phi(z)$ is a non-linear function, but the parity symmetry comes to the rescue, i.e. $X \stackrel{d}{=}-X$: $$ \mathbb{E}\left(\Phi\left(\sin(X) \cos(Y)\right)\right) = \mathbb{E}\left(\Phi\left(\sin(-X) \cos(Y)\right)\right) = \mathbb{E}\left(\Phi\left(-\sin(X) \cos(Y)\right)\right) = \mathbb{E}\left(1-\Phi\left(\sin(X) \cos(Y)\right)\right) $$ resulting in $$ \mathbb{E}\left(\Phi\left(\sin(X) \cos(Y)\right)\right) = \frac{1}{2} $$ For normals with non-zero mean $m$ and unit standard deviation here is the plot of the probability as a function of $m$, computed in Mathematica:

enter image description here

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+1. Nice symmetry argument. –  Did Oct 27 '12 at 8:52

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