STEM Problem Detail

Problem #13

Mathematics Hard

Problem Statement

Prompt

Give me the correct solution of Question 5-15 step by step and solve a,b,c,d,e,f,g,h,i,j,k.

Solution Rendering...

The joint probability density function is given as \( f(x, y) = c(x + y) \) over the region \( 0 < x < 3 \) and \( 0 < y < x \).
First, we determine the constant \( c \) by requiring that
\[
\int_{0}^{3} \int_{0}^{x} c(x + y) \, dy \, dx = 1.
\]
Evaluate the inner integral:
\[
\int_{0}^{x} (x + y) \, dy = \left[ xy + \frac{y^2}{2} \right]_{0}^{x} = x^2 + \frac{x^2}{2} = \frac{3x^2}{2}.
\]
Then
\[
\int_{0}^{3} \frac{3c x^2}{2} \, dx = \frac{3c}{2} \cdot \frac{x^3}{3} \Big|_{0}^{3} = \frac{3c}{2} \cdot 9 = \frac{27c}{2}.
\]
Setting this equal to 1 gives \( \frac{27c}{2} = 1 \), hence
\[
c = \frac{2}{27}.
\]
Therefore,
\[
f(x, y) = \frac{2}{27}(x + y), \quad 0 < x < 3, \; 0 < y < x.
\]

**(a)** \( P(X < 1, Y < 2) \).
Since the region \( 0 < x < 3, \; 0 < y < x \) restricts \( y < x \), for \( x < 1 \) we automatically have \( y < 1 \), so \( Y < 2 \) is always satisfied. Thus,
\[
P(X < 1, Y < 2) = \int_{0}^{1} \int_{0}^{x} \frac{2}{27}(x + y) \, dy \, dx.
\]
First,
\[
\int_{0}^{x} (x + y) \, dy = \frac{3x^2}{2},
\]
so
\[
\int_{0}^{1} \frac{2}{27} \cdot \frac{3x^2}{2} \, dx = \frac{1}{9} \int_{0}^{1} x^2 \, dx = \frac{1}{9} \cdot \frac{1}{3} = \frac{1}{27}.
\]
Hence,
\[
P(X < 1, Y < 2) = \frac{1}{27}.
\]

**(b)** \( P(1 < X < 2) \).
The marginal density of \( X \) is needed. Integrating over \( y \):
\[
f_X(x) = \int_{0}^{x} \frac{2}{27}(x + y) \, dy = \frac{2}{27} \cdot \frac{3x^2}{2} = \frac{x^2}{9}, \quad 0 < x < 3.
\]
Then
\[
P(1 < X < 2) = \int_{1}^{2} \frac{x^2}{9} \, dx = \frac{1}{9} \left[ \frac{x^3}{3} \right]_{1}^{2} = \frac{1}{27} (8 - 1) = \frac{7}{27}.
\]

**(c)** \( P(Y > 1) \).
We integrate the joint density over \( y > 1 \). Since \( y < x \), the condition \( y > 1 \) implies \( x > 1 \). Thus,
\[
P(Y > 1) = \int_{1}^{3} \int_{1}^{x} \frac{2}{27}(x + y) \, dy \, dx.
\]
Compute the inner integral:
\[
\int_{1}^{x} (x + y) \, dy = \left[ xy + \frac{y^2}{2} \right]_{1}^{x} = \left( x^2 + \frac{x^2}{2} \right) - \left( x + \frac{1}{2} \right) = \frac{3x^2}{2} - x - \frac{1}{2}.
\]
Then
\[
P(Y > 1) = \frac{2}{27} \int_{1}^{3} \left( \frac{3x^2}{2} - x - \frac{1}{2} \right) dx.
\]
Integrate term by term:
\[
\int_{1}^{3} \frac{3x^2}{2} \, dx = \frac{1}{2} \left[ x^3 \right]_{1}^{3} = \frac{1}{2}(27 - 1) = 13,
\]
\[
\int_{1}^{3} x \, dx = \frac{1}{2}(9 - 1) = 4,
\]
\[
\int_{1}^{3} \frac{1}{2} \, dx = \frac{1}{2}(3 - 1) = 1.
\]
Thus,
\[
\int_{1}^{3} \left( \frac{3x^2}{2} - x - \frac{1}{2} \right) dx = 13 - 4 - 1 = 8,
\]
and
\[
P(Y > 1) = \frac{2}{27} \cdot 8 = \frac{16}{27}.
\]

**(d)** \( P(X < 2, Y < 2) \).
Since \( y < x \), for \( x < 2 \) we have \( y < 2 \) automatically satisfied. However, we must only integrate over \( 0 < y < x \). Thus,
\[
P(X < 2, Y < 2) = \int_{0}^{2} \int_{0}^{x} \frac{2}{27}(x + y) \, dy \, dx.
\]
As before,
\[
\int_{0}^{x} (x + y) \, dy = \frac{3x^2}{2},
\]
so
\[
\int_{0}^{2} \frac{2}{27} \cdot \frac{3x^2}{2} \, dx = \frac{1}{9} \int_{0}^{2} x^2 \, dx = \frac{1}{9} \cdot \frac{8}{3} = \frac{8}{27}.
\]

**(e)** \( E(X) \).
\[
E(X) = \int_{0}^{3} x f_X(x) \, dx = \int_{0}^{3} x \cdot \frac{x^2}{9} \, dx = \frac{1}{9} \int_{0}^{3} x^3 \, dx = \frac{1}{9} \cdot \frac{81}{4} = \frac{9}{4}.
\]

**(f)** \( E(Y) \).
First find the marginal density of \( Y \). For fixed \( y \), \( x \) ranges from \( y \) to 3:
\[
f_Y(y) = \int_{y}^{3} \frac{2}{27}(x + y) \, dx = \frac{2}{27} \left[ \frac{x^2}{2} + yx \right]_{y}^{3}
= \frac{2}{27} \left( \frac{9}{2} + 3y - \frac{y^2}{2} - y^2 \right) = \frac{2}{27} \left( \frac{9}{2} + 3y - \frac{3y^2}{2} \right).
\]
Simplify:
\[
f_Y(y) = \frac{1}{27} \left( 9 + 6y - 3y^2 \right) = \frac{3 + 2y - y^2}{9}, \quad 0 < y < 3.
\]
Then
\[
E(Y) = \int_{0}^{3} y \cdot \frac{3 + 2y - y^2}{9} \, dy = \frac{1}{9} \int_{0}^{3} (3y + 2y^2 - y^3) \, dy.
\]
Compute:
\[
\int_{0}^{3} 3y \, dy = \frac{3}{2} \cdot 9 = \frac{27}{2},
\]
\[
\int_{0}^{3} 2y^2 \, dy = \frac{2}{3} \cdot 27 = 18,
\]
\[
\int_{0}^{3} y^3 \, dy = \frac{81}{4}.
\]
Sum:
\[
\frac{27}{2} + 18 - \frac{81}{4} = \frac{54}{4} + \frac{72}{4} - \frac{81}{4} = \frac{45}{4}.
\]
Thus,
\[
E(Y) = \frac{1}{9} \cdot \frac{45}{4} = \frac{5}{4}.
\]

**(g)** Marginal probability distribution of \( X \).
From earlier,
\[
f_X(x) = \frac{x^2}{9}, \quad 0 < x < 3.
\]

**(h)** Conditional probability distribution of \( Y \) given \( X = 1 \).
The conditional density is
\[
f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)} = \frac{\frac{2}{27}(x + y)}{\frac{x^2}{9}} = \frac{2(x+y)}{3x^2}.
\]
For \( x = 1 \):
\[
f_{Y|X=1}(y) = \frac{2(1+y)}{3} = \frac{2}{3}(1+y), \quad 0 < y < 1.
\]

**(i)** \( E(Y | X = 1) \).
\[
E(Y | X = 1) = \int_{0}^{1} y \cdot \frac{2}{3}(1+y) \, dy = \frac{2}{3} \int_{0}^{1} (y + y^2) \, dy = \frac{2}{3} \left[ \frac{y^2}{2} + \frac{y^3}{3} \right]_{0}^{1} = \frac{2}{3} \left( \frac{1}{2} + \frac{1}{3} \right) = \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9}.
\]

**(j)** \( P(Y > 2 | X = 1) \).
Since the support of \( Y \) given \( X = 1 \) is \( 0 < y < 1 \), the event \( Y > 2 \) is impossible. Hence,
\[
P(Y > 2 | X = 1) = 0.
\]

**(k)** Conditional probability distribution of \( X \) given \( Y = 2 \).
First, note that for \( Y = 2 \), the condition \( y < x \) requires \( x > 2 \), and also \( x < 3 \) from the joint support.
The conditional density is
\[
f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)}.
\]
We already have \( f_Y(y) = \frac{3 + 2y - y^2}{9} \). For \( y = 2 \):
\[
f_Y(2) = \frac{3 + 4 - 4}{9} = \frac{3}{9} = \frac{1}{3}.
\]
Thus,
\[
f_{X|Y=2}(x) = \frac{\frac{2}{27}(x + 2)}{\frac{1}{3}} = \frac{2(x+2)}{9}, \quad 2 < x < 3.
\]

The required results are:
\begin{align*}
&\text{(a) } \frac{1}{27}, \quad \text{(b) } \frac{7}{27}, \quad \text{(c) } \frac{16}{27}, \quad \text{(d) } \frac{8}{27}, \quad \text{(e) } \frac{9}{4}, \quad \text{(f) } \frac{5}{4}, \\
&\text{(g) } f_X(x) = \frac{x^2}{9}, \; 0 < x < 3, \\
&\text{(h) } f_{Y|X=1}(y) = \frac{2}{3}(1+y), \; 0 < y < 1, \\
&\text{(i) } \frac{5}{9}, \quad \text{(j) } 0, \\
&\text{(k) } f_{X|Y=2}(x) = \frac{2(x+2)}{9}, \; 2 < x < 3.
\end{align*}

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