Let $X$, $Y$, and $Z$ be three independent uniform random variables on $[0, 1]$. Compute the probability $P(XY < Z^2)$.

Here is what I've done:
\begin{align} P(XY<Z^2) &= \int_{-\infty}^{\infty} P(XY<Z^2|Z=z)f_z(z)dz \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} P(XY<z^2|Y=y)f_y(y)f_z(z)dzdy \\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} P(X<\frac{z^2}{y})f_y(y)f_z(z)dzdy\\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} F_X(\frac{z^2}{y})dzdy \end{align} since $f_y(y)=f_z(z)=1$ and $F_X(x)=x$, we get $$\int_{0}^{1}\int_{0}^{1}\frac{z^2}{y}dzdy$$ There must be something wrong...
Can anyone help me with this?

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    $\begingroup$ Hi Yibei, welcome to MSE. What have you tried so far? It is more likely to get responses if you show your efforts. $\endgroup$ – Ahmad Bazzi Sep 3 '18 at 23:26
  • $\begingroup$ I've edited my question. How to put my question back on topic? $\endgroup$ – clement Sep 4 '18 at 22:33
  • $\begingroup$ i've edited the format of your question $\endgroup$ – Ahmad Bazzi Sep 4 '18 at 22:38
  • $\begingroup$ Thank you so much! $\endgroup$ – clement Sep 4 '18 at 22:55

Nice plan, but try it while leaving $Z$ till last, since that is where the square is. This also avoids having any inverses in the probability statement.

$$\begin{split}\mathsf P(XY\leq Z^2) &=\int_0^1\int_0^1\mathsf P(Z^2\geq xy)~\mathsf d x~\mathsf d y\\ & = \int_0^1\int_0^1 (1-\sqrt{xy})~\mathsf d x~\mathsf dy\\&~~\vdots\end{split}$$

It works.

PS: Your problem was that you should have used $\mathsf P(X\leq z^2/y) = \min(z^2/y, 1)$ not just $z^2/y$. You cannot have a proability excceeding 1 which our formula would have given when $y<z^2$.

  • $\begingroup$ Thank you for your help. Changing into Z^2>xy is really a smart move! It avoids the tricky part. Thank you for your help $\endgroup$ – clement Sep 5 '18 at 20:47

Your missing the point, that $\forall y \notin [0,1] f_y(y)=0$ and for $x$ too.

I get \begin{align} P(XY<Z^2) &=\int_0^1 \int_0^1 \int_0^1 1_{xy<z^2}dx dy dz\\ &=\int_0^1 \int_0^1 \int_0^1 1_{x<z^2/y}dx dy dz\\ &=\int_0^1 \int_0^1 \int_0^{\min(z^2/y,1)}dx dy dz\\ &=\int_0^1 \int_0^1 \min(z^2/y,1) dy dz\\ &=\int_0^1 \int_0^{z^2} \min(z^2/y,1) dy + \int_{z^2}^1 \min(z^2/y,1) dy dz\\ &=\int_0^1 \int_0^{z^2} 1 dy + \int_{z^2}^1 z^2/y dy dz\\ &=\int_0^1 z^2 + z^2 (-\ln(z^2)) dz\\ &=\int_0^1 z^2(1-\ln(z^2)) dz\\ &=[(5 z^3)/9 - 1/3 z^3 \ln(z^2)]_0^1\\&=5/9 \end{align} I did the last integral with Wolframalpha

  • $\begingroup$ Could you explain more from the 5th line? Thank you! $\endgroup$ – clement Sep 5 '18 at 20:45
  • $\begingroup$ @YibeiHe do you mean from th 4th to the 5th line or from 5 to 6? $\endgroup$ – Boba Fit Sep 5 '18 at 21:03
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    $\begingroup$ Its partitioning the integration domain $[0;1]$ into $[0;z^2)\cup[z^2;1]$ because $\min$ is a piecewise function. Where $y< z^2$ there $\min(z^2/y,1)=1$. Where $y\geq z^2$, there $\min(z^2/y,1)=z^2/y$. @YibeiHe $\endgroup$ – Graham Kemp Sep 5 '18 at 21:03
  • $\begingroup$ @BobaFit Oh, I see what you mean! thank you $\endgroup$ – clement Sep 7 '18 at 0:03

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