Question
The random variables X and Y are distributed according to the joint PDF
fX,Y (x,y) = [ax², if 1 ≤ x ≤2 and 0 ≤ y ≤ x, 0, otherwise.
1 Find the constant a
a=
2 Determine the marginal PDF f Y (y)
(Your answer can be either numerical or an algebraic function of y)
(Your answer can be either numerical or an algebraic function of y).
Useful fact: You may find the following fact useful: integrate x ^ 2 d x from a to b = 1/3 * (b ^ 3 - a ^ 3) .
If 0 <= y <= 1
f_{Y}(y) =
If 1 < y <= 2
f_{Y}(y) =
3. Determine the conditional expectation of 1 / (X ^ 2 * Y) given that Y = 5/4
E[ 1/(X ^ 2 * Y) |Y= 5/4 ]=
fX,Y (x,y) = [ax², if 1 ≤ x ≤2 and 0 ≤ y ≤ x, 0, otherwise.
1 Find the constant a
a=
2 Determine the marginal PDF f Y (y)
(Your answer can be either numerical or an algebraic function of y)
(Your answer can be either numerical or an algebraic function of y).
Useful fact: You may find the following fact useful: integrate x ^ 2 d x from a to b = 1/3 * (b ^ 3 - a ^ 3) .
If 0 <= y <= 1
f_{Y}(y) =
If 1 < y <= 2
f_{Y}(y) =
3. Determine the conditional expectation of 1 / (X ^ 2 * Y) given that Y = 5/4
E[ 1/(X ^ 2 * Y) |Y= 5/4 ]=
Answers
Answered by
GPT-4o mini
To solve this problem, we will approach it step by step.
### 1. Finding the constant \( a \)
We begin by normalizing the joint PDF \( f_{X,Y}(x,y) \). The integral of the joint PDF over its defined range must equal 1:
\[
\int_{1}^{2} \int_{0}^{x} ax^2 \, dy \, dx = 1
\]
First, we calculate the inner integral:
\[
\int_{0}^{x} ax^2 \, dy = ax^2 \cdot y \bigg|_0^x = ax^2 \cdot x = ax^3
\]
Now substituting this into the outer integral:
\[
\int_{1}^{2} ax^3 \, dx
\]
Calculating the outer integral:
\[
= a \cdot \frac{1}{4} (x^4) \bigg|_1^2 = a \cdot \left(\frac{1}{4}(16 - 1)\right) = a \cdot \frac{15}{4}
\]
Setting this equal to 1 gives:
\[
a \cdot \frac{15}{4} = 1 \implies a = \frac{4}{15}
\]
### 2. Determining the marginal PDF \( f_Y(y) \)
The marginal PDF \( f_Y(y) \) is obtained by integrating the joint PDF over \( x \):
\[
f_Y(y) = \int_{1}^{2} f_{X,Y}(x,y) \, dx
\]
The joint PDF \( f_{X,Y}(x,y) = ax^2 \) is defined for \( 1 \leq x \leq 2 \) and \( 0 \leq y \leq x \).
**Case 1: \( 0 \leq y \leq 1 \)**
For this case, the valid range for \( x \) is from 1 to \( 2 \) (ensuring \( y \leq x \)):
\[
f_Y(y) = \int_{1}^{2} f_{X,Y}(x,y) \, dx = \int_{1}^{2} \frac{4}{15} x^2 \, dx
\]
Calculating the integral:
\[
= \frac{4}{15} \cdot \frac{1}{3}(x^3) \bigg|_1^2= \frac{4}{15} \cdot \frac{1}{3} (8 - 1) = \frac{4}{15} \cdot \frac{7}{3} = \frac{28}{45}
\]
Thus,
\[
f_Y(y) = \frac{28}{45}, \quad 0 \leq y \leq 1
\]
**Case 2: \( 1 < y \leq 2 \)**
For this case, the valid range for \( x \) is from \( y \) to \( 2 \):
\[
f_Y(y) = \int_{y}^{2} f_{X,Y}(x,y) \, dx = \int_{y}^{2} \frac{4}{15} x^2 \, dx
\]
Calculating this integral:
\[
= \frac{4}{15} \cdot \frac{1}{3}(x^3) \bigg|_y^2 = \frac{4}{15} \cdot \frac{1}{3}(8 - y^3) = \frac{4(8 - y^3)}{45}
\]
Thus,
\[
f_Y(y) = \frac{4(8 - y^3)}{45}, \quad 1 < y \leq 2
\]
### 3. Determine \( E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] \)
To find this conditional expectation, first find the conditional PDF \( f_{X|Y}(x|y) \):
\[
f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}
\]
Given \( Y = \frac{5}{4} \), we have \( 1 < \frac{5}{4} \leq 2 \), so we use the second case of \( f_Y(y) \) from above.
\[
f_Y\left(\frac{5}{4}\right) = \frac{4(8 - (5/4)^3)}{45}
\]
Now we compute \( f_{X,Y}(x, \frac{5}{4}) \), valid for \( \frac{5}{4} \leq x \leq 2 \):
\[
f_{X,Y}(x, \frac{5}{4}) = \frac{4}{15} x^2
\]
Thus,
\[
f_{X|Y}\left(x \mid \frac{5}{4}\right) = \frac{\frac{4}{15} x^2}{\frac{4(8 - (5/4)^3)}{45}} = \frac{15 x^2}{8 - \left(\frac{5}{4}\right)^3} = \frac{15x^2}{8 - \frac{125}{64}} = \frac{15 x^2}{\frac{512 - 125}{64}} = \frac{15 \cdot 64 x^2}{387}
\]
The expected value is:
\[
E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \int_{\frac{5}{4}}^{2} \frac{1}{x^2 \cdot \frac{5}{4}} f_{X|Y}\left(x \mid \frac{5}{4}\right) \, dx
\]
Substituting and simplifying:
\[
=\frac{4}{5} \cdot \int_{\frac{5}{4}}^{2} \frac{15 \cdot 64 x^2}{387} \cdot \frac{1}{x^2} \, dx
\]
\[
=\frac{4}{5} \cdot \frac{15 \cdot 64}{387} \cdot \int_{\frac{5}{4}}^{2} 1 \, dx
\]
The integral evaluates as:
\[
= \frac{4}{5} \cdot \frac{15 \cdot 64}{387} \cdot \left(2 - \frac{5}{4}\right) = \frac{4}{5} \cdot \frac{15 \cdot 64}{387} \cdot \frac{3}{4}
\]
This simplifies to:
\[
E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \frac{15 \cdot 64}{5 \cdot 387} = \frac{192}{387}
\]
Thus,
\[
E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \frac{64}{129}
\]
Finally, the answers are:
1. \( a = \frac{4}{15} \)
2.
- \( f_Y(y) = \frac{28}{45}, \quad 0 \leq y \leq 1 \)
- \( f_Y(y) = \frac{4(8 - y^3)}{45}, \quad 1 < y \leq 2 \)
3. \( E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \frac{64}{129} \)
### 1. Finding the constant \( a \)
We begin by normalizing the joint PDF \( f_{X,Y}(x,y) \). The integral of the joint PDF over its defined range must equal 1:
\[
\int_{1}^{2} \int_{0}^{x} ax^2 \, dy \, dx = 1
\]
First, we calculate the inner integral:
\[
\int_{0}^{x} ax^2 \, dy = ax^2 \cdot y \bigg|_0^x = ax^2 \cdot x = ax^3
\]
Now substituting this into the outer integral:
\[
\int_{1}^{2} ax^3 \, dx
\]
Calculating the outer integral:
\[
= a \cdot \frac{1}{4} (x^4) \bigg|_1^2 = a \cdot \left(\frac{1}{4}(16 - 1)\right) = a \cdot \frac{15}{4}
\]
Setting this equal to 1 gives:
\[
a \cdot \frac{15}{4} = 1 \implies a = \frac{4}{15}
\]
### 2. Determining the marginal PDF \( f_Y(y) \)
The marginal PDF \( f_Y(y) \) is obtained by integrating the joint PDF over \( x \):
\[
f_Y(y) = \int_{1}^{2} f_{X,Y}(x,y) \, dx
\]
The joint PDF \( f_{X,Y}(x,y) = ax^2 \) is defined for \( 1 \leq x \leq 2 \) and \( 0 \leq y \leq x \).
**Case 1: \( 0 \leq y \leq 1 \)**
For this case, the valid range for \( x \) is from 1 to \( 2 \) (ensuring \( y \leq x \)):
\[
f_Y(y) = \int_{1}^{2} f_{X,Y}(x,y) \, dx = \int_{1}^{2} \frac{4}{15} x^2 \, dx
\]
Calculating the integral:
\[
= \frac{4}{15} \cdot \frac{1}{3}(x^3) \bigg|_1^2= \frac{4}{15} \cdot \frac{1}{3} (8 - 1) = \frac{4}{15} \cdot \frac{7}{3} = \frac{28}{45}
\]
Thus,
\[
f_Y(y) = \frac{28}{45}, \quad 0 \leq y \leq 1
\]
**Case 2: \( 1 < y \leq 2 \)**
For this case, the valid range for \( x \) is from \( y \) to \( 2 \):
\[
f_Y(y) = \int_{y}^{2} f_{X,Y}(x,y) \, dx = \int_{y}^{2} \frac{4}{15} x^2 \, dx
\]
Calculating this integral:
\[
= \frac{4}{15} \cdot \frac{1}{3}(x^3) \bigg|_y^2 = \frac{4}{15} \cdot \frac{1}{3}(8 - y^3) = \frac{4(8 - y^3)}{45}
\]
Thus,
\[
f_Y(y) = \frac{4(8 - y^3)}{45}, \quad 1 < y \leq 2
\]
### 3. Determine \( E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] \)
To find this conditional expectation, first find the conditional PDF \( f_{X|Y}(x|y) \):
\[
f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}
\]
Given \( Y = \frac{5}{4} \), we have \( 1 < \frac{5}{4} \leq 2 \), so we use the second case of \( f_Y(y) \) from above.
\[
f_Y\left(\frac{5}{4}\right) = \frac{4(8 - (5/4)^3)}{45}
\]
Now we compute \( f_{X,Y}(x, \frac{5}{4}) \), valid for \( \frac{5}{4} \leq x \leq 2 \):
\[
f_{X,Y}(x, \frac{5}{4}) = \frac{4}{15} x^2
\]
Thus,
\[
f_{X|Y}\left(x \mid \frac{5}{4}\right) = \frac{\frac{4}{15} x^2}{\frac{4(8 - (5/4)^3)}{45}} = \frac{15 x^2}{8 - \left(\frac{5}{4}\right)^3} = \frac{15x^2}{8 - \frac{125}{64}} = \frac{15 x^2}{\frac{512 - 125}{64}} = \frac{15 \cdot 64 x^2}{387}
\]
The expected value is:
\[
E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \int_{\frac{5}{4}}^{2} \frac{1}{x^2 \cdot \frac{5}{4}} f_{X|Y}\left(x \mid \frac{5}{4}\right) \, dx
\]
Substituting and simplifying:
\[
=\frac{4}{5} \cdot \int_{\frac{5}{4}}^{2} \frac{15 \cdot 64 x^2}{387} \cdot \frac{1}{x^2} \, dx
\]
\[
=\frac{4}{5} \cdot \frac{15 \cdot 64}{387} \cdot \int_{\frac{5}{4}}^{2} 1 \, dx
\]
The integral evaluates as:
\[
= \frac{4}{5} \cdot \frac{15 \cdot 64}{387} \cdot \left(2 - \frac{5}{4}\right) = \frac{4}{5} \cdot \frac{15 \cdot 64}{387} \cdot \frac{3}{4}
\]
This simplifies to:
\[
E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \frac{15 \cdot 64}{5 \cdot 387} = \frac{192}{387}
\]
Thus,
\[
E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \frac{64}{129}
\]
Finally, the answers are:
1. \( a = \frac{4}{15} \)
2.
- \( f_Y(y) = \frac{28}{45}, \quad 0 \leq y \leq 1 \)
- \( f_Y(y) = \frac{4(8 - y^3)}{45}, \quad 1 < y \leq 2 \)
3. \( E\left[\frac{1}{X^2 Y} \mid Y = \frac{5}{4}\right] = \frac{64}{129} \)