What is the probability that $P(\frac{xy}{z}\le t)$? This is a problem I do not know how to solve. Assume $x,y,z$ are uniformly identically distributed variables at $[0,1]$. I am supposed to calculate $$P\left(\frac{xy}{z}\le t\right)$$
I thought the computation would be similar to the case $P(xy\le t)=t-t\log[t]$, since we have $$P\left(\frac{xy}{z}\le t\right)=\int \limits^{1}_{0}P(xy\le zt) \, dz$$So since I do not know $zt$'s value in principle, I can calculate it by $$\int \limits^{1}_{0}(zt-zt\log[zt]) \, dz$$
But this turned out to be very different from the answer in the solution manual, where the author distinguished cases if $t\le 1$ or $t> 1$, and the two answers are remarkablely different(My answer only works when $0<t\le 1$, but his answer is $1-\frac{t}{4}$ regardless of $z$ when $0<t\le 1$). So I want to ask for a hint. Sorry the problem is really of low level.
 A: Given $t>0$, you want to calculate the volume of the set $S=(0,1)^3\cap\bigl\{(x,y,z)\in\mathbb R^3: xy\leq tz\bigr\}$, which can be done using iterated integration. First: what are are the possible values of $x$ for $(x,y,z)\in S$? You have $0<x<1$ as a necessary condition. Conversely: given $x_0\in(0,1)$, can you find $y,z\in(0,1)$ such that $x_0y\leq tz$? the answer is affirmative: take any $z$, say $z=1/2$, and now take $y$ small so that $2x_0y\leq t$. Therefore your volume can be written as the following iterated integral (with incomplete data):
$$\int_{x=0}^{x=1}\int_{y=?}^{y=?}\int_{z=?}^{z=?}1\,dz\,dy\,dx\,.$$
Now, given $x_0\in(0,1)$, what are the possible values of $y$ for points $(x_0,y,z)\in S$? you must have $y\leq tz/x_0<t/x_0$, and so $0<y<\min\{1,t/x_0\}$ is a necessary condition. Conversely, if $y$ satisfies this condition, then you can take $z$ very near to $1$ in such manner that $y<(t/x_0)z$. Thus, our integral becomes
$$\int_{x=0}^{x=1}\int_{y=0}^{y=\min\{1,t/x\}}\int_{z=?}^{z=?}1\,dz\,dy\,dx\,.$$
Finally, given $x_0,y_0\in(0,1)$ such that $x_0y_0<t$, what are the possible values of $z$ for points $(x_0,y_0,z)\in S$? We have $0<z<1$ and $z\geq x_0y_0/t$ as a necessary condition, and so we have $x_0y_0/t\leq z<1$, which clearly is also a sufficient condition. Thus, the final form of our desired integral is
$$\begin{align*}
\int_{x=0}^{x=1}\int_{y=0}^{y=\min\{1,t/x\}}\int_{z=\frac{xy}t}^{z=1}1\,dz\,dy\,dx=&\,\int_{x=0}^{x=1}\int_{y=0}^{y=\min\{1,t/x\}}1-\frac{xy}t\,dy\,dx\\
=&\,\int_{x=0}^{x=1}\biggl[y-\frac{xy^2}{2t}\biggr]\Biggl|_{y=0}^{y=\min\{1,t/x\}}\,dx\\
=&\,\int_{x=0}^{x=1}\min\{1,t/x\}-\frac x{2t}\,\bigl(\min\{1,t/x\}\bigr)^2\,dx\,.
\end{align*}$$
Finally, you must treat the cases $t<1$ and $t\geq1$ separately: in the former case, you split the integration interval, obtaining
$$\begin{align*}
&\,\int_{x=0}^{x=t}\min\{1,t/x\}-\frac x{2t}\,\bigl(\min\{1,t/x\}\bigr)^2\,dx\\
+&\,\int_{x=t}^{x=1}\min\{1,t/x\}-\frac x{2t}\,\bigl(\min\{1,t/x\}\bigr)^2\,dx\\[2mm]
=&\,\int_{x=0}^{x=t}1-\frac x{2t}\,dx+\,\int_{x=t}^{x=1}\frac tx-\frac x{2t}\biggl(\frac tx\biggr)^2\,dx\\[2mm]
=&\,\biggl(x-\frac{x^2}{4t}\biggr)\Biggl|_{x=0}^{x=t}+\biggl(\frac t2\,\log(x)\biggr)\Biggl|_{x=t}^{x=1}\\[2mm]
=&\,\frac{3t}4-\frac{t\log(t)}2\,;
\end{align*}$$
in the latter case you have $\min\{1,t/x\}=1$ for all $x\in(0,1)$, to the integral becomes
$$\int_{x=0}^{x=1}1-\frac x{2t}\,dx=1-\frac1{4t}\,.$$
A: I try to redo the calculation here. The base case $P(xy\le t)$ can be calculated by:


*

*$0<t<1$: $t-t\log[t]$

*$t>1$: $1$

*$t=0:0$


So to calculate $$\int^{1}_{0}P(xy<zt)dz$$ I have to split up in 3 cases. Either $zt=0$, which either $z=0$ or $t=0$. The first case can be ignored; the second one has probability $0$. If $0<zt<1$, then we have $z<\frac{1}{t}$. So if $t>1$ then we split the integral as 
$$\int^{\frac{1}{t}}_{0}P(xy<zt)dz+\int^{1}_{\frac{1}{t}}P(xy<zt)dz=t\int^{\frac{1}{t}}_{0}(z-z[\log[z]+\log[t]])dz+(1-\frac{1}{t})$$
I am still double checking the integral. 
