Consider the independent random variables $U_1 =U(0,c)$ $U_2=U(0,\pi/2)$

The challenge is to compute the probability:

$$P(U_1 \leq a\text{cos}(U_2))$$

An attempt:

$$P(U_1 \leq a\text{cos}(U_2))=P(U_1-acos(U_2)\leq0)$$

Applying convolution: $$\int_{-\infty}^0\int_{-\infty}^{\infty} f_{U_1}(t)f_{-a\text{cos}(U_2)}(x-t)dtdx$$ By independence: $$=\int_{-\infty}^0\int_{-\infty}^{\infty} f_{U_1,-a\text{cos}(U_2)}(t,x-t)dtdx$$

This is where I get stuck, I believe using multivariate change of variable does the trick. The transform should give:

$$=\int_0^{\pi/2}\int_0^{a \text{cos(t)}} f_{U_1,U_2}(x,t)dxdt$$

From which it's no trouble to carry out the computation to the end. Is it correct to use multivariable change of variable in the above, and if so, what transformation should be performed?

  • 2
    $\begingroup$ \begin{align} P(U_1\le a\cos U_2)&=\frac{2}{\pi c}\iint_{x\le a\cos y}\mathbf1_{0<x<c,0<y<\pi/2}\,dx\,dy \\&=\frac{2}{\pi c}\int_0^{\pi/2}\int_0^{\min(c,a\cos y)}\,dx\,dy \end{align} $\endgroup$ – StubbornAtom Jan 21 at 6:54
  • 2
    $\begingroup$ You can just start directly with:$$P(U_1\leq a\cos U_2)=\mathbb E\mathbf1_{U_1\leq a\cos U_2}=\int\int\mathbf1_{u\leq a\cos v}f_{U,V}(u,v)dudv$$ leading to the expression mentioned by @StubbornAtom. No change of variables (or other tricks) is required. $\endgroup$ – drhab Jan 21 at 9:54
  • $\begingroup$ @drhab I can see that the method is quite favorable. But CAN this be solved using the transform? $\endgroup$ – Dole Jan 22 at 0:03
  • $\begingroup$ Why do you think that $P(U_1-a\cos U_2\leq0)=P(U_1+a\cos U_2\leq0)$? $\endgroup$ – drhab Jan 22 at 10:11
  • $\begingroup$ @drhab Good catch, corrected! $\endgroup$ – Dole Jan 22 at 11:14

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