Let $X$ be a continuous random variable with uniform probability distribution between $0$ and $S$, i.e, $X \sim U(0, S)$. Let Y be another continuous random variable distributed uniformly between $0$ and $X$, i.e, $Y \sim U(0, X)$.

  1. I want to know if there is a valid joint probability distribution.
  2. If my direction for caclulating the joint CDF as shown below is correct.

$F_{X,Y}(u_{o}, v_{o}) = P(X<u_{o}, Y<v_{o}) = \frac{P(Y<v_{o} | X < u_{o}) P(X < u_{o}) }{P(Y < v_{o})} $

$P(Y<v_{o} | X < u_{o}) = \int_{-\infty}^{u_{o}}{P(Y<v_{o} | X = u)du}$

$P(X < u_{o}) = \frac{u_{o}}{S}$

$P(Y < v_{o}) = \int_{-\infty}^{\infty}{P(Y < v_{o} | X = u)du}$

  • $\begingroup$ Why using the notations $u_0,v_0$ instead of the common $x,y$? $\endgroup$
    – drhab
    Jan 14, 2020 at 11:18
  • $\begingroup$ Yeah, my bad. I was reading a text with these notations and they just stuck. $\endgroup$
    – kunalc92
    Jan 14, 2020 at 11:44

1 Answer 1


It is evident that $Y\leq X$ almost surely.

Consequently if $u_{0}\leq v_{0}$ then: $$P\left(X\leq u_{0},Y\leq v_{0}\right)=P\left(X\leq u_{0}\right)$$ and the RHS is easy to find.

If $0<v_{0}<u_{0}<S$ then:

$$\begin{aligned}P\left(X\leq u_{0},Y\leq v_{0}\right) & =P\left(X\leq v_{0},Y\leq v_{0}\right)+P\left(v_{0}<X\leq u_{0},Y\leq v_{0}\right)\\ & =P\left(X\leq v_{0}\right)+\int_{v_{0}}^{u_{0}}P\left(Y\leq v_{0}\mid X=x\right)f_{X}\left(x\right)dx\\ & =\frac{v_{0}}{S}+\frac{1}{S}\int_{v_{0}}^{u_{0}}P\left(Y\leq v_{0}\mid X=x\right)dx\\ & =\frac{v_{0}}{S}+\frac{1}{S}\int_{v_{0}}^{u_{0}}\frac{v_{0}}{x}dx\\ & =\frac{v_{0}}{S}+\frac{v_{0}}{S}\left[\ln x\right]_{v_{0}}^{u_{0}}\\ & =\frac{v_{0}}{S}+\frac{v_{0}}{S}\left(\ln u_{0}-\ln v_{0}\right) \end{aligned} $$

Personally I would go for the notation: $$P\left(X\leq x,Y\leq y\right)=\frac{y}{S}+\frac{y}{S}\left(\ln x-\ln y\right)$$where $0<y<x<S$.

  • $\begingroup$ Thanks @drhab. Could you please point out what is wrong in my approach other than not taking the probability density inside the integral ? $\endgroup$
    – kunalc92
    Jan 14, 2020 at 11:53
  • $\begingroup$ Also, what would be the correct way to formulate $P(X < x | Y = y)$. Is this zero since Y is continuous ? $\endgroup$
    – kunalc92
    Jan 14, 2020 at 12:21
  • 1
    $\begingroup$ $F_{X,Y}(u_0,v_0)=P(Y\leq v_0\mid X_0\leq u_0)P(X_0\leq u_0)$, so dividing by $P(Y<v_0)$ (as you do) is incorrect. Also you give expressions but did not really calculate them. The question in your second comment is not relevant so it is not appropriate to work that out in a comment here. You could pose it as a new question. If the condition ($Y=y$ here) has probability $0$ then that does not mean that the conditional probability equals $0$. It does mean that we must be careful because the common definition $P(A\mid B)=P(A\cap B)/P(B)$ cannot be practicized, since $P(B)=0$ is not allowed then. $\endgroup$
    – drhab
    Jan 14, 2020 at 12:22
  • 1
    $\begingroup$ Is there a typo in the second line of the second equation: "$=P(X\le u_0)$"? $\endgroup$
    – user140541
    Jan 14, 2020 at 15:59
  • $\begingroup$ @d.k.o. Yes, there is. Thank you for attending me. I have repaired. Also the following lines were actually wrong because of that typo. $\endgroup$
    – drhab
    Jan 14, 2020 at 16:19

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