Let, X,Y are two independent RVs following $U(0,1)$. Let, $W=XI_{\{Y\leq X^2\}}$, where $I_{A}$ denotes the indicator function on the set $A$. Find out the CDF(Cumulative Distribution Function) of W.


$P(W\leq w)=P(X\leq w, Y\leq X^2)+P(Y\geq X^2)=P(\sqrt{Y}\leq X \leq w )+P(X^2 \leq Y)$

Now, $P(\sqrt{Y}\leq X \leq w )=\int_{0}^{1}(\int_{\sqrt{y}}^{w} 1 dx) 1 dy=w-\frac{2}{3}$.

Again, $P(X^2 \leq Y)=\int_{0}^{1}(\int_{0}^{\sqrt{y}} 1 dx) 1 dy=\frac{2}{3}$.

Therefore, $P(W\leq w)=w$.

Is it correct?

  • 1
    $\begingroup$ Could you explain the reasoning behind your answer? In particular, why do you add in $P(Y\ge X^2)$? $\endgroup$ – whuber Aug 23 '18 at 18:54
  • $\begingroup$ Best to write the second term as $$\mathbb{P}(W \leq w, Y \geq X^2)=\mathbb{P}(0 \leq w, Y \geq X^2)=\boldsymbol 1_{\{w\geq0\}}\mathbb{P}(Y \geq X^2) $$ $\endgroup$ – Rohit Arora Aug 23 '18 at 22:16

Your calculation of $P(X\leq w, Y\leq X^2)=P(\sqrt{Y}\leq X \leq w )$ is incorrect; it returns a negative probability when $w=0$. You can write the event $\{X\leq w, Y\leq X^2\}$ as $\{(X,Y)\in A\}$ where $A$ is the region in the $(x,y)$ plane bounded by $y=x^2$, $y=0$ and $x=w$. Visualizing the region $A$ will give you the right limits for the double integral: $$P (X\leq w, Y\leq X^2)=\iint_A1\,dy\,dx=\text{Area of $A$}=\int_{x=0}^w\left(\int_{y=0}^{x^2}1\,dy\right)dx.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.