Answer :
Alright, let’s break down this problem step-by-step:
### (a) Finding the constant \( k \)
Given the probability density function (PDF) for \( X \):
[tex]\[ f_X(x) = \begin{cases} k(10-x) & \text{if } 0 \leq x \leq 10 \\ 0 & \text{otherwise} \end{cases} \][/tex]
We need to find the constant \( k \). To do this, we use the property that the total probability over the entire range of \( x \) must equal 1. Hence, we integrate \( f_X(x) \) over its domain and set the integral equal to 1:
[tex]\[ \int_{0}^{10} k(10-x) \, dx = 1 \][/tex]
Now, compute the integral:
[tex]\[ k \int_{0}^{10} (10 - x) \, dx = 1 \][/tex]
We can break this into two simpler integrals:
[tex]\[ k \left[ \int_{0}^{10} 10 \, dx - \int_{0}^{10} x \, dx \right] = 1 \][/tex]
Evaluate each integral separately:
[tex]\[ \int_{0}^{10} 10 \, dx = 10x \big|_0^{10} = 100 \][/tex]
[tex]\[ \int_{0}^{10} x \, dx = \frac{x^2}{2} \big|_0^{10} = \frac{100}{2} = 50 \][/tex]
Therefore, we have:
[tex]\[ k (100 - 50) = 1 \implies k \cdot 50 = 1 \implies k = \frac{1}{50} \][/tex]
So, the constant \( k \) is:
[tex]\[ k = \frac{1}{50} \][/tex]
### (b) Finding the cumulative distribution function (CDF) and the expected value of \( W \)
The random variable \( W \) is defined by \( W = \min\{X, 5\} \). To find the cumulative distribution function (CDF) \( F_W(w) \), we consider the following cases:
1. Case \( w < 0 \): [tex]\[ F_W(w) = 0 \][/tex]
2. Case \( 0 \leq w < 5 \): Here, \( W \) takes the value \( w \) only if \( X \leq w \). Thus,
[tex]\[ F_W(w) = P(W \leq w) = P(X \leq w) = \int_{0}^{w} f_X(x) \, dx = \int_{0}^{w} \frac{1}{50}(10 - x) \, dx \][/tex]
Evaluate the integral:
[tex]\[ F_W(w) = \frac{1}{50} \left[ 10w - \frac{w^2}{2} \right] \][/tex]
3. Case \( w \geq 5 \): Here, we need to consider two parts:
- For \( 0 \leq x < 5 \), \( W = X \)
- For \( x \geq 5 \), \( W = 5 \)
[tex]\[ F_W(w) = P(W \leq w) = P(X \leq 5) + P(5 \leq w \leq X) \][/tex]
But since \( W \leq 5 \) always when \( X \) is greater or equal to 5:
[tex]\[ F_W(w) = P(X \leq 5) = \int_{0}^{5} \frac{1}{50}(10 - x) \, dx \][/tex]
Evaluate the integral:
[tex]\[ \int_{0}^{5} \frac{1}{50}(10 - x) \, dx = \frac{1}{50} \left[ 10x - \frac{x^2}{2} \right] \bigg|_0^{5} = \frac{1}{50} (50 - 12.5) = \frac{37.5}{50} = 0.75 \][/tex]
So, for \( w \geq 5 \):
[tex]\[ F_W(w) = 0.75 \][/tex]
Combining all our results, the CDF \( F_W(w) \) is:
[tex]\[ F_W(w) = \begin{cases} 0 & \text{if } w < 0 \\ \frac{1}{50} \left[ 10w - \frac{w^2}{2} \right] & \text{if } 0 \leq w < 5 \\ 0.75 & \text{if } w \geq 5 \end{cases} \][/tex]
### Expected Value of \( W \)
The expected value \( E[W] \) can be found by splitting it into two parts:
1. \( E[X \ | \ X \leq 5] \)
2. \( 5 \cdot P(X > 5) \)
First, compute \( E[X \ | \ X \leq 5] \):
[tex]\[ E[X \ | \ X \leq 5] = \int_{0}^{5} x f_X(x) \, dx = \int_{0}^{5} x \frac{1}{50}(10 - x) \, dx \][/tex]
Evaluate the integral:
[tex]\[ \int_{0}^{5} x \frac{1}{50}(10 - x) \, dx = \frac{1}{50} \int_{0}^{5} (10x - x^2) \, dx \][/tex]
[tex]\[ = \frac{1}{50} \left[ 5x^2 - \frac{x^3}{3} \right] \bigg|_0^{5} = \frac{1}{50} \left[ 5(25) - \frac{125}{3} \right] = \frac{1}{50} \left[ 125 - \frac{125}{3} \right] \][/tex]
[tex]\[ = \frac{1}{50} \left[ \frac{375 - 125}{3} \right] = \frac{1}{50} \left[ \frac{250}{3} \right] = \frac{5}{3} \][/tex]
Now, \( E[5 \ | \ X > 5] \):
[tex]\[ E[5 \ | \ X > 5] = 5 \cdot P(X > 5) = 5 \cdot \left( 1 - P(X \leq 5) \right) = 5 \cdot \left( 1 - 0.75 \right) = 5 \cdot (0.25) = 1.25 \][/tex]
Now combine these two results:
[tex]\[ E[W] = \frac{5}{3} + 1.25 = \frac{5}{3} + \frac{5}{4} = \frac{20}{12} + \frac{15}{12} = \frac{35}{12} \approx 2.92 \][/tex]
The expected value is:
[tex]\[ E[W] \approx 2.92 \][/tex]
Thus, we have:
- \( k = \frac{1}{50} \)
- CDF \( F_W(w) \):
[tex]\[ F_W(w) = \begin{cases} 0 & \text{if } w < 0 \\ \frac{1}{50} \left[ 10w - \frac{w^2}{2} \right] & \text{if } 0 \leq w < 5 \\ 0.75 & \text{if } w \geq 5 \end{cases} \][/tex]
- Expected value [tex]\( E[W] \approx 2.92 \)[/tex]
### (a) Finding the constant \( k \)
Given the probability density function (PDF) for \( X \):
[tex]\[ f_X(x) = \begin{cases} k(10-x) & \text{if } 0 \leq x \leq 10 \\ 0 & \text{otherwise} \end{cases} \][/tex]
We need to find the constant \( k \). To do this, we use the property that the total probability over the entire range of \( x \) must equal 1. Hence, we integrate \( f_X(x) \) over its domain and set the integral equal to 1:
[tex]\[ \int_{0}^{10} k(10-x) \, dx = 1 \][/tex]
Now, compute the integral:
[tex]\[ k \int_{0}^{10} (10 - x) \, dx = 1 \][/tex]
We can break this into two simpler integrals:
[tex]\[ k \left[ \int_{0}^{10} 10 \, dx - \int_{0}^{10} x \, dx \right] = 1 \][/tex]
Evaluate each integral separately:
[tex]\[ \int_{0}^{10} 10 \, dx = 10x \big|_0^{10} = 100 \][/tex]
[tex]\[ \int_{0}^{10} x \, dx = \frac{x^2}{2} \big|_0^{10} = \frac{100}{2} = 50 \][/tex]
Therefore, we have:
[tex]\[ k (100 - 50) = 1 \implies k \cdot 50 = 1 \implies k = \frac{1}{50} \][/tex]
So, the constant \( k \) is:
[tex]\[ k = \frac{1}{50} \][/tex]
### (b) Finding the cumulative distribution function (CDF) and the expected value of \( W \)
The random variable \( W \) is defined by \( W = \min\{X, 5\} \). To find the cumulative distribution function (CDF) \( F_W(w) \), we consider the following cases:
1. Case \( w < 0 \): [tex]\[ F_W(w) = 0 \][/tex]
2. Case \( 0 \leq w < 5 \): Here, \( W \) takes the value \( w \) only if \( X \leq w \). Thus,
[tex]\[ F_W(w) = P(W \leq w) = P(X \leq w) = \int_{0}^{w} f_X(x) \, dx = \int_{0}^{w} \frac{1}{50}(10 - x) \, dx \][/tex]
Evaluate the integral:
[tex]\[ F_W(w) = \frac{1}{50} \left[ 10w - \frac{w^2}{2} \right] \][/tex]
3. Case \( w \geq 5 \): Here, we need to consider two parts:
- For \( 0 \leq x < 5 \), \( W = X \)
- For \( x \geq 5 \), \( W = 5 \)
[tex]\[ F_W(w) = P(W \leq w) = P(X \leq 5) + P(5 \leq w \leq X) \][/tex]
But since \( W \leq 5 \) always when \( X \) is greater or equal to 5:
[tex]\[ F_W(w) = P(X \leq 5) = \int_{0}^{5} \frac{1}{50}(10 - x) \, dx \][/tex]
Evaluate the integral:
[tex]\[ \int_{0}^{5} \frac{1}{50}(10 - x) \, dx = \frac{1}{50} \left[ 10x - \frac{x^2}{2} \right] \bigg|_0^{5} = \frac{1}{50} (50 - 12.5) = \frac{37.5}{50} = 0.75 \][/tex]
So, for \( w \geq 5 \):
[tex]\[ F_W(w) = 0.75 \][/tex]
Combining all our results, the CDF \( F_W(w) \) is:
[tex]\[ F_W(w) = \begin{cases} 0 & \text{if } w < 0 \\ \frac{1}{50} \left[ 10w - \frac{w^2}{2} \right] & \text{if } 0 \leq w < 5 \\ 0.75 & \text{if } w \geq 5 \end{cases} \][/tex]
### Expected Value of \( W \)
The expected value \( E[W] \) can be found by splitting it into two parts:
1. \( E[X \ | \ X \leq 5] \)
2. \( 5 \cdot P(X > 5) \)
First, compute \( E[X \ | \ X \leq 5] \):
[tex]\[ E[X \ | \ X \leq 5] = \int_{0}^{5} x f_X(x) \, dx = \int_{0}^{5} x \frac{1}{50}(10 - x) \, dx \][/tex]
Evaluate the integral:
[tex]\[ \int_{0}^{5} x \frac{1}{50}(10 - x) \, dx = \frac{1}{50} \int_{0}^{5} (10x - x^2) \, dx \][/tex]
[tex]\[ = \frac{1}{50} \left[ 5x^2 - \frac{x^3}{3} \right] \bigg|_0^{5} = \frac{1}{50} \left[ 5(25) - \frac{125}{3} \right] = \frac{1}{50} \left[ 125 - \frac{125}{3} \right] \][/tex]
[tex]\[ = \frac{1}{50} \left[ \frac{375 - 125}{3} \right] = \frac{1}{50} \left[ \frac{250}{3} \right] = \frac{5}{3} \][/tex]
Now, \( E[5 \ | \ X > 5] \):
[tex]\[ E[5 \ | \ X > 5] = 5 \cdot P(X > 5) = 5 \cdot \left( 1 - P(X \leq 5) \right) = 5 \cdot \left( 1 - 0.75 \right) = 5 \cdot (0.25) = 1.25 \][/tex]
Now combine these two results:
[tex]\[ E[W] = \frac{5}{3} + 1.25 = \frac{5}{3} + \frac{5}{4} = \frac{20}{12} + \frac{15}{12} = \frac{35}{12} \approx 2.92 \][/tex]
The expected value is:
[tex]\[ E[W] \approx 2.92 \][/tex]
Thus, we have:
- \( k = \frac{1}{50} \)
- CDF \( F_W(w) \):
[tex]\[ F_W(w) = \begin{cases} 0 & \text{if } w < 0 \\ \frac{1}{50} \left[ 10w - \frac{w^2}{2} \right] & \text{if } 0 \leq w < 5 \\ 0.75 & \text{if } w \geq 5 \end{cases} \][/tex]
- Expected value [tex]\( E[W] \approx 2.92 \)[/tex]
