Practice set 5.3

1

Suppose \(X\) and \(Y\) are jointly continuous with joint pdf given by:

\[ f(x, y) = \begin{cases} 3x, & 0 \le y \le x \le 1\\ 0, & \text{elsewhere.} \end{cases} \]

A

Find the marginal density functions of \(X\) and \(Y\). Do either of them follow well-known forms?

\[f_X(x) = \int_{0}^{x} 3x\,dy = 3x \cdot x = 3x^2, \qquad 0 \le x \le 1.\]

So \(X\sim BETA(3,1)\).

\[f_Y(y) = \int_{y}^{1} 3x\,dx = \frac{3}{2}(1 - y^2), \qquad 0 \le y \le 1.\]

Not quite a well-known form!

B

For what values of \(x\) is \(f_{Y|X}(y|x)\) defined? Find this conditional density, and identify it as following a well-known form.

For \(x \in (0,1)\):

\[f_{X|Y}(x|y)=\frac{f(x,y)}{f_X(x)} = \frac{3x}{3x^2} = \begin{cases}\frac{1}{x} & 0 \leq y \leq x\\ 0 & otherwise \end{cases}\]

Thus \(Y|X=x \sim UNIF(0,x)\).

C

Find \(F_{Y|X}(y|x)\).

Using the properties of uniforms,

\[F_{Y|X}(y|x) = \begin{cases} 0 & y < 0 \\ y/x & 0 \le y \le x \\ 1 & y > x \end{cases}\]

D

What is \(P(Y>1/2 | X = 3/4)\)?

\[P(Y>1/2 | X = 3/4) = 1-F_{X|Y}(0.5|0.75) = 1-(1/2)/(3/4) = \frac{1}{3}\]

E

What is \(P(Y<1/2 | X = 1/4)\)?

\[P(Y<1/2 | X = 1/4) = F_{X|Y}(1/2|1/4) = 1\]

2

Let X and Y have joint density function:

\[f_{X,Y}(x, y) = \begin{cases} e^{-(x + y)}, & x > 0, \; y > 0, \\ 0, & \text{elsewhere.} \end{cases}\]

with joint CDF:

\[F_{X,Y}(x,y) = \begin{cases} 0, & x \le 0 \text{ or } y \le 0, \\ (1 - e^{-x})(1 - e^{-y}), & x > 0, y > 0 \end{cases}\]

Find the marginal CDFs of \(X\) and \(Y\). Identify \(X\) and \(Y\) as marginally following well-known forms.

\[F_{X}(x) = \lim_{y\rightarrow \infty} F_{X,Y}(x,y) = \begin{cases} 0 & x \leq 0 \\ 1-e^{-x} & x > 0 \end{cases}\]

\[F_{Y}(y) = \lim_{x\rightarrow \infty} F_{X,Y}(x,y) = \begin{cases} 0 & y \leq 0 \\ 1-e^{-y} & y > 0 \end{cases}\]

Marginally, both \(X\) and \(Y\) are \(EXP(1)\) random variables as they each have the \(EXP(1)\) CDF.

3

\(X\) and \(Y\) are jointly continuous with joint pdf:

\[f_{X,Y}(x,y) = \begin{cases}6(1-y) & 0 < x < y < 1 \\ 0 & otherwise \end{cases}\]

and joint CDF:

\[F_{X,Y}(x,y) = \begin{cases} 0 & x < 0\ or\ y < 0 \\ 3x^2-2x^3 + 3x(y-x)(2-y-x) & 0 < x < y < 1 \\ 3y^2-2y^3 & 0 < y < 1, x \geq y \\ x^3-3x^2+3x & 0 < x < 1, y\geq 1\\ 1 & x\geq 1, y \geq 1 \end{cases}\]

A

Find \(F_{Y|X}(y|x)\).

We need first the marginal density of \(X\).

Note that

\[F_X(x) = \lim_{y\rightarrow \infty}F_{X,Y}(x,y) = x^3-3x^2-3x\] \[\implies f_X(x) = \frac{d}{dx}F_X(x) = 3x^2-6x-3, 0 < x < 1\]

Thus, for \(x \in (0,1)\):

\[f_{Y|X}(y|x) = \frac{6(1-y)}{3x^2-6x+3} = \frac{2(1-y)}{(x-1)^2}, x < y < 1\]

For \(x < y < 1\):

\[F_{Y|X}(y|x) = \int_x^y\frac{2(1-t)}{(x-1)^2}\ dt = \frac{1}{(x-1)^2}(2t-t^2)\big|_x^y\]

\[ = \frac{2y-y^2-2x+x^2}{(x-1)^2}\]

\[F_{Y|X}(y|x) = \begin{cases} 0 &y \le x\\ \frac{2y-y^2-2x+x^2}{(x-1)^2} & x < y < 1\\ 1 & y \geq 1 \end{cases}\]

B

Find \(F_{X|Y}(x|y)\).

We need first the marginal density of \(Y\).

Note that

\[F_Y(y) = \lim_{x\rightarrow \infty}F_{X,Y}(x,y) =3y^2-2y^3\] \[\implies f_Y(y) = 6y(1-y), 0 < y < 1\]

Thus, for \(y \in (0,1)\):

\[f_{X|Y}(x|y) = \frac{6(1-y)}{6y(1-y)} = \frac{1}{y}, 0 < x < y\]

Note that \(X|Y=y \sim UNIF(0, 1/y)\)!

Thus:

\[F_{X|Y}(x|y) = \begin{cases} 0 & x < 0 \\ \frac{x}{y} & 0 \leq x \leq y \\ 1 & x > y\end{cases}\]

4

Suppose that the random variables X and Y have joint probability density function given by

\[f(x, y) = \begin{cases} 6x^2y, & 0 \le x \le y,\; x + y \le 2, \\ 0, & \text{elsewhere.} \end{cases}\]

A

Find the marginal pdfs of \(X\) and \(Y\). Do either of them follow well-known forms?

\[f_X(x) = \int_{x}^{2 - x} 6x^2 y \, dy= 6x^2 \left[ \frac{y^2}{2} \right]_{y=x}^{y=2-x} = 3x^2\left( (2-x)^2 - x^2 \right) = 3x^2(4 - 4x)\]

\[= 12x^2 - 12x^3= 12x^2(1-x), 0 < x < 1\]

So \(X\sim BETA(3,2)\).

\[f_Y(y) = \int_{0}^{y} 6x^2 y \, dx= 6y \left[ \frac{x^3}{3} \right]_{0}^{y} = 6y \cdot \frac{y^3}{3} = 2y^4, \qquad 0 \le y \le 1.\]

So \(Y\sim BETA(5,1)\).

B

For what values of \(x\) does \(f_{Y|X}(y|x)\) exist? Find this conditional density.

For \(x\in (0,1)\):

\[f_{Y \mid X}(y \mid x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{6x^2 y}{12x^2(1 - x)} = \frac{y}{2(1 - x)},\ x \le y \le 2 - x\]

C

Find \(F_{Y|X}(y|x)\).

\[F_{Y\mid X}(y\mid x) = \int_{x}^{y} f_{Y\mid X}(t\mid x)\,dt = \begin{cases} 0, & y < x,\\[6pt] \dfrac{y^2-x^2}{4(1-x)}, & x \le y \le 2-x,\\[8pt] 1, & y > 2-x. \end{cases}\]

D

Find \(P(Y<1.1 | X = 0.6)\).

\[P(Y<1.1\mid X=0.6)=F_{Y\mid X}(1.1\mid 0.6) = \frac{1.1^2 - 0.6^2}{4(1-0.6)}= 0.53125.\]

E

Find \(P(Y>1.5 | X = 0.8)\).

\[P(Y>1.5\mid X=0.8)=1-F_{Y\mid X}(1.5\mid 0.8) = 1-1 = 0,\] since \(1.5 > 2-0.8 = 1.2\).

F

Find \(P(Y > 1.5 | X = 1.6)\).

This is \(1-F_{X|Y}(1.5|1.6) = 1-0 = 1\) since we have \(y < x\).

5

Suppose \(X\) and \(Y\) are uniformly distributed over the triangular region shown below:

A

Find the marginal density functions.

Recall we found previously:

\[f(x,y) = \begin{cases} 1 & y-1 \leq x \leq 1-y, 0 \leq y \leq 1 \\ 0 & otherwise \end{cases}\]

\[f_Y(y) = \int_{y-1}^{1-y} 1\ dx = (y-1)-(1-y) = 2(y-1),\ 0\le y \le 1\]

When \(x \in (-1,0):\)

\[f_X(x) = \int_0^{x+1} 1\ dy = x+1\]

When \(x \in (0,1):\)

\[f_X(x) = \int_0^{1-x} 1\ dy = 1-x\]

Expressed more succinctly:

\[f_X(x) = 1-|x|,\ -1 \leq x \leq 1\]

B

What is \(P(Y>1/2 | X = 1/4)\)?

For \(x = 1/4\), we have \(f_X(1/4) = 1-1/4\). Thus:

\[f_{Y|X}(y|1/4) = \frac{1}{1-1/4} = \frac{4}{3}, 0 \le y \le \frac{3}{4}\]

I.e., \(Y|X=1/4 \sim UNIF(0, 3/4)\). Using properties of uniforms, \(P(Y>1/2 | X=1/4) = \frac{1}{3}\).

6

If \(X\sim UNIF(0,1)\) and for \(0<x<1\), \(Y|X=x \sim UNIF(0,x)\):

A

Find the joint pdf of \(X\) and \(Y\).

\[f(x,y) = f_{Y|X}(y|x) \cdot f_X(x) = \frac{1}{x}\cdot 1 = \begin{cases} \frac{1}{x} & 0 < y \leq x \le 1 \\ 0 & otherwise \\ \end{cases}\]

B

Find the marginal density function for \(Y\).

\[f_Y(y) = \int_y^1 \frac{1}{x}\ dx = \ln(x)\big|_y^1 = \begin{cases} -\ln(y) & 0 < y < 1\\ 0 & otherwise \end{cases}\]

7

If \(X\sim N(0,1)\) and \(Y|X=x \sim N(x, 1)\), show that marginally \(Y\sim N(0,2)\).

First step, find \(f(x,y)\).

\[f(x,y) = f_{Y|X}(y|x) \cdot f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-(y-x)^2/2}\cdot \frac{1}{\sqrt{2\pi}}e^{-x^2/2} = \frac{1}{2\pi}e^{-(y^2-2yx+x^2)/2}e^{-x^2/2}\] \[= \frac{1}{2\pi}e^{-(y^2-2yx+2x^2)/2}, y \in \mathbb{R}, x \in \mathbb{R}\]

Now for the marginal:

\[f_Y(y) = \int_{-\infty}^\infty \frac{1}{2\pi}e^{-(y^2-2yx+2x^2)/2}\ dx\] \[= \frac{1}{2\pi}e^{-y^2/2}\int_{-\infty}^\infty e^{-(2x^2-2xy)/2}\ dx\] \[=\frac{1}{2\pi}e^{-y^2/2}\int_{-\infty}^\infty e^{-(x^2-xy)/2\cdot (0.5)}\ dx\] Complete the square:

\[=\frac{1}{2\pi}e^{-y^2/2}\int_{-\infty}^\infty e^{-(x^2-xy+y^2/4)/2\cdot (0.5)}e^{y^2/4}\ dx\]

\[=\frac{1}{2\pi}e^{-y^2/2}e^{y^2/4}\int_{-\infty}^\infty e^{-(x-y/2)^2/2\cdot (0.5)}\ dx\]

\[= \frac{1}{2\pi}e^{-y^2/4}\cdot \sqrt{2\pi \cdot 0.5},\]

noting the integrand as a kernel of a \(N(y/2, 0.5)\) density. Simplifying:

\[f_Y(y) = \frac{1}{\sqrt{2\pi\cdot 2}}e^{-y^2/(2\cdot 2)}, y\in \mathbb{R}\]

\(\therefore Y\sim N(0,2)\).