Practice set 5.2

For all, sketch the regions of integration!!!

1

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

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

A

What is \(P(X - Y > 0.5)\)?

Because this density has a constant height, we can just recognize this as the volume of a triangular wedge with depth and width of base both equal to 0.5 and the height of the wedge equal to 1. The volume would be \(\frac{1}{2^3} = \frac{1}{8}\).

Or integrate: \[P(X - Y > 0.5) = \int_{0.5}^{1} \int_{0}^{x-0.5} 1 \, dy \, dx = \int_{0.5}^{1} (x - 0.5) \, dx\]

\[= \left[\frac{x^2}{2} - 0.5x\right]_{0.5}^{1} = \left(\frac{1}{2} - \frac{1}{2}\right) - \left(\frac{1}{8} - \frac{1}{4}\right) = 0 - \left(-\frac{1}{8}\right) = \frac{1}{8}\]

B

What is \(P(XY < 0.5)\)?

\[P(XY < 0.5) = 1-P(XY>0.5) = \int_{0.5}^1\int_{0.5/x}^1 1 \, dy \, dx\] \[\begin{aligned} &= \Big[ x \Big]_{0.5}^{1} - \frac{1}{2} \Big[ \ln|x| \Big]_{0.5}^{1}\\ &= (1 - 0.5) - \frac{1}{2} (\ln 1 - \ln 0.5) \\ &= 0.5 - \frac{1}{2} (0 - (-\ln 2)) \\ &= 0.5 - \frac{\ln 2}{2}\\ \end{aligned}\]

2

Let X and Y have joint density function

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

A

Find the joint CDF.

For \(x>0, y>0\): \[ F_{X,Y}(x,y) = \int_0^x \int_0^y e^{-(u+v)} \, dv \, du\] \[\int_0^y e^{-(u+v)} \, dv = e^{-u} \int_0^y e^{-v} \, dv = e^{-u} \left[ - e^{-v} \right]_0^y = e^{-u} (1 - e^{-y}) \]

\[\int_0^x e^{-u} (1 - e^{-y}) \, du = (1 - e^{-y}) \int_0^x e^{-u} du = (1 - e^{-y}) \left[ - e^{-u} \right]_0^x= (1 - e^{-y}) (1 - e^{-x})\]

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

B

What is \(P(X < 1, \, Y > 5)\)?

\[P(X < 1, Y > 5) = \int_0^1 \int_5^\infty e^{-(x+y)} \, dy \, dx\]

Step 1: Integrate with respect to \(y\):

\[\int_5^\infty e^{-(x+y)} \, dy = e^{-x} \int_5^\infty e^{-y} \, dy = e^{-x} \left[ - e^{-y} \right]_5^\infty = e^{-x} e^{-5} = e^{-(x+5)}\]

Step 2: Integrate with respect to x:

\[\int_0^1 e^{-(x+5)} dx = e^{-5} \int_0^1 e^{-x} dx = e^{-5} \left[ - e^{-x} \right]_0^1 = e^{-5} (1 - e^{-1})\]

\[\boxed{P(X < 1, Y > 5) = e^{-5}(1 - e^{-1})}\]

C

What is \(P(X + Y < 3)\)?

\[P(X + Y < 3) = \int_0^3 \int_0^{3-x} e^{-(x+y)} \, dy \, dx\]

Step 1: Integrate with respect to \(y\):

\[\int_0^{3-x} e^{-(x+y)} \, dy = e^{-x} \int_0^{3-x} e^{-y} dy = e^{-x} (1 - e^{-(3-x)}) = e^{-x} - e^{-3}\]

Step 2: Integrate with respect to \(x\):

\[\int_0^3 (e^{-x} - e^{-3}) dx = \int_0^3 e^{-x} dx - \int_0^3 e^{-3} dx\] \[= \left[ - e^{-x} \right]_0^3 - 3 e^{-3} = (1 - e^{-3}) - 3 e^{-3} = 1 - 4 e^{-3}\]

\[\boxed{P(X + Y < 3) = 1 - 4 e^{-3}}\]

3

Let X and Y have the joint probability density function given by

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

A

Find the value of \(k\) that makes this a probability density function.

\[\int_0^1 \int_0^1 kxy \, dy \, dx = 1 \implies k \int_0^1 x \left[ \frac{y^2}{2} \right]_0^1 dx = k \int_0^1 \frac{x}{2} dx = k \frac{1}{4} = 1 \implies k = 4\]

B

Find the joint distribution function for X and Y.

For \((x,y)\in Support\):

\[F_{X,Y}(x,y) = \int_0^x \int_0^y 4 u v \, dv \, du = \int_0^x 4 u \frac{v^2}{2}\Big|_0^y du = \int_0^x 2 u y^2 du = y^2 x^2\]

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

C

Find \(P(X \le 1/2, \, Y \le 3/4)\).

\[P(X \le 1/2, Y \le 3/4) = F_{X,Y}(1/2, 3/4) = (1/2)^2 (3/4)^2 = 9/64\]

D

Find \(P(X \le 3/4, Y < 1.5)\).

\[P(X \le 3/4, Y < 1.5) = F_{X,Y}(3/4, 1.5) = (3/4)^2 = 9/16\]

4

Let X and Y have the joint probability density function given by

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

A

Find the value of \(k\) that makes this a probability density function.

\[\int_0^1 \int_0^1 k(1-y) \, dx \, dy = k \int_0^1 (1-y) dx dy = k \int_0^1 (1-y) dy = k \left[ y - \frac{y^2}{2} \right]_0^1 = k (1/2) = 1 \implies k = 2\]

B

Find \(P(X<Y)\).

\[P(X<Y) = \int_0^1 \int_0^y 2(1-y) \, dx \, dy =\int_0^1\left( 2(1-y) \cdot x \Big|_0^y\right)dy = \int_0^12y(1-y)dy\] \[= 2 \int_0^1 (y - y^2) \, dy = 2 \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_0^1\] \[= 2 \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{3}\]

5

Let X and Y denote the proportions of two different types of components in a sample from a mixture of chemicals used as an insecticide. Suppose that X and Y have the joint density function given by

\[ f(x, y) = \begin{cases} 2, & 0 \le x \le 1,\; 0 \le y \le 1,\; 0 \le x + y \le 1, \\ 0, & \text{elsewhere.} \end{cases} \]

(Notice that \(X + Y \le 1\) because the random variables denote proportions within the same sample.)

A

Find the joint cumulative distribution function.

Use geometry for this one! Draw the regions!

For \((x,y)\) in the support, CDF is the volume of a rectangular cube with height = 2, base dimensions of \(x\) and \(y\). \(F_{X,Y}(x,y) = 2xy\).

If \(x+y>1\) but \(0\leq x \leq 1\) and \(0 \leq y \leq 1\), subtract the volume of two triangular bases with areas of \(\frac{(1-x)^2}{2}\) and \(\frac{(1-y)^2}{2}\). \(F_{X,Y}(x,y) = 1-(1-x)^2-(1-y)^2\)

Find the other two cases by letting \(y=1\) or \(x=1\).

Full CDF:

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

B

Find \(P(X \le 3/4,\, Y \le 3/4).\)

\[F(3/4, 3/4) = 1-(1/4)^2 - (1/4)^2 = 14/16\]

C

Find \(P(X \le 1/2,\, Y \le 1/4).\)

\[F(1/2, 1/2) = 2\cdot 1/2 \cdot 1/4 = 1/4\]

D

Find \(P(X\leq 3/4, Y \leq 2).\)

\[F(3/4, 2) = 1-1/4^2 = 15/16\]

6

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

Verify that this is a valid joint density function.

Want: \[ \int_{0}^{1}\int_{x}^{2-x} 6x^{2}y \, dy\,dx.\]

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

\[\int_{0}^{1} 3x^{2}(4 - 4x)\,dx = 12\int_{0}^{1} (x^{2} - x^{3})\,dx = 12\left[\frac{x^{3}}{3} - \frac{x^{4}}{4}\right]_{0}^{1} = 12\left(\frac{1}{3} - \frac{1}{4}\right) = 1.\]

B

What is the probability that \(X + Y\) is less than 1?

\[P(X + Y < 1) = \int_{0}^{1/2} \int_{x}^{\,1-x} 6x^{2}y \, dy \, dx.\]

\[ \int_{x}^{1-x} 6x^{2}y \, dy = 6x^{2}\left[\frac{y^{2}}{2}\right]_{y=x}^{1-x} = 3x^{2}\left((1-x)^{2} - x^{2}\right)\]

\[(1-x)^{2} - x^{2} = (1 - 2x + x^{2}) - x^{2} = 1 - 2x.\] \[\Rightarrow 3x^{2}(1 - 2x) = 3x^{2} - 6x^{3}.\]

\[\int_{0}^{1/2} (3x^{2} - 6x^{3})\,dx = \left[ x^{3} - \frac{3}{2}x^{4} \right]_{0}^{1/2} = \frac{1}{8} - \frac{3}{2}\cdot\frac{1}{16} = \frac{1}{8} - \frac{3}{32} = \frac{1}{32}.\]

7

The management at a fast-food outlet is interested in the joint behavior of the random variables \(X\) and \(Y\), defined as the total time between a customer’s arrival at the drive-thru and departure from the service window, and the time a customer waits in line before reaching the service window, respectively.

Because \(Y\) includes the time a customer waits in line, we must have \(Y \le X\).

The relative frequency distribution of observed values of \(X\) and \(Y\) can be modeled by the probability density function

\[ f(x, y) = \begin{cases} e^{-x}, & 0 \le y \le x < \infty, \\ 0, & \text{elsewhere.} \end{cases} \]

with time measured in minutes.

A

Find the joint CDF.

For \((x,y)\) in the support:

\[F_{X,Y}(x,y) =\int_{0}^{y} \int_{u}^{x} e^{-v}\, dv\, du\]

\[\int_{u}^{x} e^{-v}\, dv = \left[-e^{-v}\right]_{v=u}^{v=x} = e^{-u} - e^{-x}.\]

\[\int_{0}^{y} \left(e^{-u} - e^{-x}\right)\, du = \int_{0}^{y} e^{-u}\, du - \int_{0}^{y} e^{-x}\, du= 1-e^{-y}-ye^{-x}\]

\[F_{X,Y}(x,y) = \begin{cases} 0 & x < 0 \mbox{ or } y < 0\\ 1-e^{-y}-ye^{-x} & 0 < y \leq x \\ 1-e^{-x}-xe^{-x} & 0 < x < y\\ \end{cases}\]

B

Use the joint CDF to find \(P(X < 2,\, Y > 1)\)

Draw the region to note that:

\[P(X<2, Y>1) = P(X<2,Y<2)-P(X<2,Y<1) = F_{X,Y}(2,2)-F_{X,Y}(2,1)\] \[ = (1-e^{-2}-2e^{-2})-(1-e^{-1}-e^{-2}) = e^{-1}-2e^{-2}\]

C

Use the joint CDF to find \(P(X>2, Y < 1)\).

Draw the region to note that;

\[P(X>2, Y < 1) = P(X<\infty, Y < 1) - P(X<2,Y<1) = F_{X,Y}(\infty, 1)-F_{X,Y}(2,1)\]

\[ = 1-e^{-1} -(1-e^{-1}-e^{-2}) = e^{-2}\]

D

Find \(P(X>2, Y>1)\).

Draw the region to note that:

\[P(X>2, Y>1) = 1-(P(X\leq 2, Y \leq 1) + P(X\leq 2, Y > 1) + P(X \geq 2, Y \leq 1) )\]

\[ = 1-(1-e^{-1}-e^{-2} + e^{-1}-2e^{-2} + e^{-2}) = 2e^{-2}\]

E

Find \(P(X \ge 2Y)\)

\[P(X\ge 2Y) = \int_{x=0}^{\infty}\int_{y=0}^{x/2} e^{-x}\,dy\,dx = \int_{0}^{\infty} \left(\frac{x}{2}\right)e^{-x}\,dx = \frac{1}{2}\int_{0}^{\infty} x e^{-x}\,dx.\]

Note the integral is \(\Gamma(2) =1\).

Thus \(P(X\ge 2Y) = \frac{1}{2}\).

8

Let \((X, Y)\) denote the coordinates of a point chosen at random inside a unit circle whose center is at the origin.

That is, \(X\) and \(Y\) have a joint density function given by

\[ f(x, y) = \begin{cases} \dfrac{1}{\pi}, & x^2 + y^2 \le 1, \\ 0, & \text{elsewhere.} \end{cases} \]

Find \(P(X \le Y).\)

By symmetry, the line \(y = x\) divides the unit circle into two equal halves. The region where \(x \le y\) is exactly half the circle.

Answer: \(\boxed{\frac{1}{2}}\)

9

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

A

What is the joint probability density function?

The height is constant, given \(X\) and \(Y\) are uniformly distributed. The area of the region is the area of two triangles, each with base = height = 1. So the two triangles have area 1/2 each thus the total area of the region is 1, hence the height of the joint pdf is 1 as well. It remains to define the limits of the support.

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

B

Find \(P(Y>|X|)\).

Draw the region to note that \(P(Y>|X|)\) is the volume of a cube with square base of area \((\sqrt{2}/2)^2\) and height = 1. Thus \(P(Y>|X|) = 1/2\).