Joint density of $X = \min(U_{1}, U_{2})$ and $Y = \max(U_{2}, U_{3})$

Question

Let $U_{1}, U_{2}, U_{3}$ be i.i.d. random variables which are uniformly distributed on (0,1). Let $X = \min(U_{1}, U_{2})$ and $Y = \max(U_{2}, U_{3})$

(a) Find $P(X< x, Y < y)$ for all x,y in $\mathbb R$

(b) Find $P(X = Y)$

(c) Find $E[XI_{\{X=Y\}}]$ where $I_{A}$ is the indicator function of A.

(a) I found the PDFs of X and Y, i.e. $f_X(x)=2(1-x), f_Y(y)=2y$ using order statistics. I wanted to find the joint density and then integrate it to get the joint CDF, but I'm not sure how to find the joint density given that X and Y are dependent (?). Can someone please help?

b) $$P(X=Y)=P( \min(U_1,U_2)= \max(U_2,U_3))=P( \min(U_1,U_2)=U_2, \max(U_2,U_3)=U_2)=P(U_1>U_2>U_3)=\frac{1}{6}$$ (Since there are 6 (3!) ways to order 3 of these variables, I'm neglecting the possibility of $U_1=U_2=U_3$). Is this correct?

c) Isn't it the same as part (b)? I'm confused, please help.

Answer 1

For finding the joint CDF, it's probably easiest to attack it directly and break it into components rather than trying to sort out the joint PDF from the marginal distributions.

$\begin{eqnarray} P(X < x, Y < y) & = & P(X < x | Y < y) P(Y < y) \\ P(X < x | Y < y) & = & 1 - P(X \geq x | Y < y) \\ & = & 1 - P(U_1 \geq x \land U_2 \geq x | U_2 < y \land U_3 < y) \\ & = & 1 - P(U_1 \geq x) P(U_2 \geq x | U_2 < y) & \textrm{by independence} \\ P(Y < y) & = & P(U_2 < y \land U_3 < y) \\ & = & y^2 \end{eqnarray}$

I'll leave it to you to piece everything together from there, and maybe do some calculus to prove to yourself that you get the right marginal distributions as a result. You can then apply the joint CDF to parts (b) and (c) as necessary.

Answer 2

(a). The joint distribution $F(x, y) := P(X < x, Y < y)$ clearly depends on the value of $(x, y)$:

• If $x > 1$ and $y > 1$, then clearly $F(x, y) = 1$.
• If $x \leq 0$ or $y \leq 0$, then clearly $F(x, y) = 0$.
• If $0 < x \leq 1$ and $y > 1$ then $$F(x, y) = P(X < x) = 1 - P(\min(U_1, U_2) \geq x) = 1 - (1 - x)^2 = 2x - x^2.$$
• If $x > 1$ and $0 < y \leq 1$, then $$F(x, y) = P(Y < y) = P(U_2 < y)P(U_3 < y) = y^2. $$

The only non-trivial case is when $0 < x \leq 1$ and $0 < y \leq 1$, for which by the independence assumption: \begin{align} & P(X < x, Y < y) = P(\min(U_1, U_2) < x, \max(U_2, U_3) < y) \\ =& \int_0^1 P(\min(U_1, z) < x, \max(z, U_3) < y)dz \tag{1} \\ =& \int_0^1 [P(\max(z, U_3) < y) - P(\min(U_1, z) \geq x, \max(z, U_3) < y)]dz \\ =& \int_0^1 P(\max(z, U_3) < y)dz - \int_0^1P(\min(U_1, z) \geq x, \max(z, U_3) < y)dz \\ =& \int_0^y P(U_3 < y) dz - \int_0^1P(\min(U_1, z) \geq x, \max(z, U_3) < y)dz \\ =:& y^2 - I \end{align}

For the integral $I$, if $x \geq y$, then the integrand is $0$ whence $I = 0, F(x, y) = y^2$. If $x < y$, then \begin{align*} I = \int_0^1P(\min(U_1, z) \geq x, \max(z, U_3) < y)]dz = \int_x^y P(U_1 \geq x, U_3 < y) dz = (1 - x)y(y - x). \end{align*} It then follows that $F(x, y) = y^2 - y(1 - x)(y - x) = xy + xy^2 - x^2y$. For the veracity of equation $(1)$, refer to Theorem $20.3$ of Probability and Measure by P. Billingsley.

The above calculations can then be summarized as \begin{align} F(x, y) = \begin{cases} 0 & (x, y) \in \{(u, v): u \leq 0\} \cup \{(u, v): v \leq 0\}, \\ xy + xy^2 - x^2y & (x, y) \in \{(u, v): 0 < u < v < 1\}, \\ y^2 & (x, y) \in \{(u, v): 0 < v \leq u < 1\} \cup \{(u, v): u > 1, 0 < v \leq 1\}, \\ 2x - x^2 & (x, y) \in \{(u, v): 0 < u \leq 1, v > 1\}, \\ 1 & (x, y) \in \{(u, v): u > 1, v > 1\}. \end{cases} \end{align}

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(c). (c) is clearly different from (b), which is identical to $E[I_{\{X = Y\}}]$ (i.e., without multiplying $X$). To evaluate $E[XI_{\{X = Y\}}]$, note that it's the expectation of a non-negative random variable, therefore, \begin{align} E[XI_{\{X = Y\}}] = \int_0^\infty P[XI_{\{X = Y\}} > t] dt = \int_0^1 P[XI_{\{X = Y\}} > t] dt = \int_0^1 P[X > t, X = Y] dt. \tag{2} \end{align} The probability in the integrand can be evaluated in the same manner as determining $F(x, y)$, for which we apply the Theorem leading to $(1)$ again:
\begin{align*} & P[X > t, X = Y] = P[\min(U_1, U_2) > t, \min(U_1, U_2) = \max(U_2, U_3)] \\ =& \int_0^1 P[\min(U_1, s) > t, U_1 \geq s, U_3 \leq s] ds \\ =& \int_t^1 P[U_1 > t, U_1 \geq s, U_3 \leq s] ds \\ =& \int_t^1 P[U_1 \geq s, U_3 \leq s]ds \\ =& \int_t^1 s(1 - s)ds = \frac{1}{6} - \frac{1}{2}t^2 + \frac{1}{3}t^3. \end{align*}

Substituting it into $(2)$, it then follows that \begin{align*} E[XI_{\{X = Y\}}] = \int_0^1 \left(\frac{1}{6} - \frac{1}{2}t^2 + \frac{1}{3}t^3 \right) dt = \frac{1}{12}. \end{align*}