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

I used the following approach :

Step 1 : Calculated the Probability distribution for $XY$. It turns out to be $P(XY \leq K) = \frac 1K$.

Step 2 : Calculated the Probability distribution for $Z^2$. It turns out to be $P(Z^2 \leq L) = P(-\root \of{L} \leq Z \leq \root \of{L}) = \root \of{L}$.

Step 3: Calculate the joint density function by multiplying the above functions and differentiating. I get $$f_{Q_1Q_2} (q_1,q_2) = \frac {-1}{2q_1^2\root\of{q_2}}$$. where $Q_1 = XY$ , $Q_2 =Z^2$.

Step 4: Calculate the probability using the integral below. $$\int_{0}^{1}\int_{q_1}^{1}\frac{-1}{2q_1^2\root\of{q_2}}dq_2dq_1$$

There is definitely something wrong with this procedure. Any help will be appreciated.

SmarthBansal
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3 Answers3

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Alternative (avoiding calculation of CDF's and PDF's): $$\begin{aligned}P\left(XY\leq Z^{2}\right) & =\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left[xy\leq z^{2}\right]dzdxdy\\ & =\int_{0}^{1}\int_{0}^{1}1-x^{\frac{1}{2}}y^{\frac{1}{2}}dxdy\\ & =1-\int_{0}^{1}\int_{0}^{1}x^{\frac{1}{2}}y^{\frac{1}{2}}dxdy\\ & =1-\int_{0}^{1}y^{\frac{1}{2}}\int_{0}^{1}x^{\frac{1}{2}}dxdy\\ & =1-\int_{0}^{1}x^{\frac{1}{2}}dx\times\int_{0}^{1}y^{\frac{1}{2}}xdy\\ & =1-\left(\int_{0}^{1}x^{\frac{1}{2}}dx\right)^{2}\\ & =1-\left(\left[\frac{2}{3}x^{\frac{3}{2}}\right]_{0}^{1}\right)^{2}\\ & =1-\left(\frac{2}{3}\right)^{2}\\ & =\frac{5}{9} \end{aligned} $$

Here $\left[xy\leq z^{2}\right]$ denotes the function $[0,1]^3\to\mathbb R$ that takes value $1$ if $xy\leq z^{2}$ and takes value $0$ otherwise.

drhab
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  • +1 Wow, I think I'm new to this. How do you get the equality in the first step? :P – SmarthBansal Sep 29 '19 at 11:28
  • The first step is based on $P(XY\leq Z^2)=\mathbb E\mathbf1_{XY\leq Z^2}$ – drhab Sep 29 '19 at 11:40
  • To make things more clear: if $f(x,y,z)=1$ if $xy\leq z^2$ and $f(x,y,z)=0$ otherwise then $P(XY\leq Z^2)=\mathbb Ef(X,Y,Z)$. That is applied here. – drhab Sep 29 '19 at 11:46
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I think the CDF of $XY$ is not correct.

\begin{align} P(XY \le K) &= 1-P(XY >K)\\ &= 1- \int_{K}^1\int_{\frac{K}{x}}^1 \, dy\, dx \\ &= 1-\int_K^1 (1-\frac{K}{x}) \, dx\\ &= 1-[x-K\ln x]_K^1 \\ &= 1-(1-K+K\ln K) \\ &= K - K \ln K \end{align}

Note that density can't be negative in step $3$.

$$f_{Q_1,Q_2}(q_1, q_2)=\frac{-\ln q_1}{2\sqrt{q_2}}$$

\begin{align}&\int_0^1\int_{q_1}^1 -\frac{\ln q_1}{2\sqrt{q_2}}\, dq_2\, dq_1 = \frac59 \end{align}

Edit to check your integral:

\begin{align} P(XY \le K) &= \int_K^1 \int_0^{\frac{K}{x}}\, dy \, dx + \int_0^K\int_0^1 \, dy \, dx\\ &=\int_K^1 \frac{K}{x}\, dx + K\\ &=K\ln x|_K^1+K\\ &=K-K\ln K \end{align}

Siong Thye Goh
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from Does the square of uniform distribution have density function? we know that $Z\sim U[0,1] => f_{Z^2}(t) = \frac{1}{2\sqrt{t}}$

and from here product distribution of two uniform distribution, what about 3 or more we know that $X\sim U[0,1],Y\sim U[0,1]\,iid => F_{XY}(t) = t-t\log t$

from Finding probability $P(X<Y)$ we know that $P(X<Y) = \int F_X(t) f_Y(t)dt$

$$P(XY<Z^2) = \int_0^1P(XY<t)f_{Z^2}(t)dt = \int_0^1(t - t\log(t))\frac{1}{2\sqrt{t}}dt = \frac{1}{2}\int_0^1 \sqrt{t} - \sqrt{t}\log t\,dt = (\frac{5}{9} t^{3/2} - t^{3/2} \log t) |_0^1 = \frac{5}{9}$$

quester
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