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To find the probability density function (pdf) of [tex]\( U = Y_1 - Y_2 \)[/tex] and subsequently its expected value [tex]\( E(U) \)[/tex], we'll proceed through several steps.
### Step 1: Range of [tex]\( U \)[/tex]
First, we need to determine the range of [tex]\( U \)[/tex]. Given that [tex]\( 0 \leq Y_2 \leq Y_1 \leq 1 \)[/tex]:
[tex]\[ U = Y_1 - Y_2 \][/tex]
Given [tex]\( Y_1 \)[/tex] ranges from 0 to 1 and [tex]\( Y_2 \)[/tex] from 0 to [tex]\( Y_1 \)[/tex]:
[tex]\[ 0 \leq U \leq 1 \][/tex]
### Step 2: Determine the Joint Pdf of [tex]\( Y_1 \)[/tex] and [tex]\( Y_2 \)[/tex]
The joint pdf of [tex]\( Y_1 \)[/tex] and [tex]\( Y_2 \)[/tex] is:
[tex]\[ f(y_1, y_2) = \left\{ \begin{array}{ll} 3 y_1 & \text{if } 0 \leq y_2 \leq y_1 \leq 1, \\ 0 & \text{elsewhere} \end{array} \right. \][/tex]
### Step 3: Transformation to New Variable [tex]\( U \)[/tex]
We need to express [tex]\( f_{U}(u) \)[/tex] in terms of the new variable [tex]\( U \)[/tex].
Define [tex]\( U = Y_1 - Y_2 \)[/tex]. Thus, [tex]\( Y_2 = Y_1 - U \)[/tex].
For a given [tex]\( u \)[/tex], [tex]\( y_2 = y_1 - u \)[/tex]. Consequently, [tex]\( y_1 \)[/tex] ranges from [tex]\( u \)[/tex] to 1, since [tex]\( y_1 ≥ y_2 \)[/tex]:
[tex]\[ \int_u^1 f(y_1, y_1 - u) dy_1 \][/tex]
### Step 4: Calculate the Pdf of [tex]\( U \)[/tex]
Inserting [tex]\( y_2 = y_1 - u \)[/tex] into [tex]\( f(y_1, y_2) \)[/tex]:
[tex]\[ f(y_1, y_1 - u) = 3 y_1 \][/tex]
So,
[tex]\[ f_U(u) = \int_u^1 3 y_1 dy_1 \][/tex]
### Step 5: Solve the Integral
Performing the integration:
[tex]\[ f_U(u) = \int_u^1 3 y_1 dy_1 = \left[ \frac{3 y_1^2}{2} \right]_u^1 = \frac{3}{2} \left(1^2 - u^2\right) = \frac{3}{2} (1 - u^2) \][/tex]
Hence the probability density function of [tex]\( U \)[/tex] is:
[tex]\[ f_U(u) = \left\{ \begin{array}{ll} \frac{3}{2} (1 - u^2) & \text{if } 0 \leq u \leq 1, \\ 0 & \text{elsewhere} \end{array} \right. \][/tex]
### Step 6: Expected Value [tex]\( E(U) \)[/tex]
Finally, let's calculate [tex]\( E(U) \)[/tex]:
[tex]\[ E(U) = \int_0^1 u f_U(u) du = \int_0^1 u \frac{3}{2} (1 - u^2) du \][/tex]
Simplifying the integral:
[tex]\[ E(U) = \frac{3}{2} \int_0^1 \left( u - u^3 \right) du = \frac{3}{2} \left( \int_0^1 u du - \int_0^1 u^3 du \right) \][/tex]
Evaluating the integrals:
[tex]\[ \int_0^1 u du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2} \][/tex]
[tex]\[ \int_0^1 u^3 du = \left[ \frac{u^4}{4} \right]_0^1 = \frac{1}{4} \][/tex]
Combining these:
[tex]\[ E(U) = \frac{3}{2} \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{3}{2} \times \frac{1}{4} = \frac{3}{8} \][/tex]
Thus, the expected value [tex]\( E(U) \)[/tex] is [tex]\( \frac{3}{8} \)[/tex].
### Step 1: Range of [tex]\( U \)[/tex]
First, we need to determine the range of [tex]\( U \)[/tex]. Given that [tex]\( 0 \leq Y_2 \leq Y_1 \leq 1 \)[/tex]:
[tex]\[ U = Y_1 - Y_2 \][/tex]
Given [tex]\( Y_1 \)[/tex] ranges from 0 to 1 and [tex]\( Y_2 \)[/tex] from 0 to [tex]\( Y_1 \)[/tex]:
[tex]\[ 0 \leq U \leq 1 \][/tex]
### Step 2: Determine the Joint Pdf of [tex]\( Y_1 \)[/tex] and [tex]\( Y_2 \)[/tex]
The joint pdf of [tex]\( Y_1 \)[/tex] and [tex]\( Y_2 \)[/tex] is:
[tex]\[ f(y_1, y_2) = \left\{ \begin{array}{ll} 3 y_1 & \text{if } 0 \leq y_2 \leq y_1 \leq 1, \\ 0 & \text{elsewhere} \end{array} \right. \][/tex]
### Step 3: Transformation to New Variable [tex]\( U \)[/tex]
We need to express [tex]\( f_{U}(u) \)[/tex] in terms of the new variable [tex]\( U \)[/tex].
Define [tex]\( U = Y_1 - Y_2 \)[/tex]. Thus, [tex]\( Y_2 = Y_1 - U \)[/tex].
For a given [tex]\( u \)[/tex], [tex]\( y_2 = y_1 - u \)[/tex]. Consequently, [tex]\( y_1 \)[/tex] ranges from [tex]\( u \)[/tex] to 1, since [tex]\( y_1 ≥ y_2 \)[/tex]:
[tex]\[ \int_u^1 f(y_1, y_1 - u) dy_1 \][/tex]
### Step 4: Calculate the Pdf of [tex]\( U \)[/tex]
Inserting [tex]\( y_2 = y_1 - u \)[/tex] into [tex]\( f(y_1, y_2) \)[/tex]:
[tex]\[ f(y_1, y_1 - u) = 3 y_1 \][/tex]
So,
[tex]\[ f_U(u) = \int_u^1 3 y_1 dy_1 \][/tex]
### Step 5: Solve the Integral
Performing the integration:
[tex]\[ f_U(u) = \int_u^1 3 y_1 dy_1 = \left[ \frac{3 y_1^2}{2} \right]_u^1 = \frac{3}{2} \left(1^2 - u^2\right) = \frac{3}{2} (1 - u^2) \][/tex]
Hence the probability density function of [tex]\( U \)[/tex] is:
[tex]\[ f_U(u) = \left\{ \begin{array}{ll} \frac{3}{2} (1 - u^2) & \text{if } 0 \leq u \leq 1, \\ 0 & \text{elsewhere} \end{array} \right. \][/tex]
### Step 6: Expected Value [tex]\( E(U) \)[/tex]
Finally, let's calculate [tex]\( E(U) \)[/tex]:
[tex]\[ E(U) = \int_0^1 u f_U(u) du = \int_0^1 u \frac{3}{2} (1 - u^2) du \][/tex]
Simplifying the integral:
[tex]\[ E(U) = \frac{3}{2} \int_0^1 \left( u - u^3 \right) du = \frac{3}{2} \left( \int_0^1 u du - \int_0^1 u^3 du \right) \][/tex]
Evaluating the integrals:
[tex]\[ \int_0^1 u du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2} \][/tex]
[tex]\[ \int_0^1 u^3 du = \left[ \frac{u^4}{4} \right]_0^1 = \frac{1}{4} \][/tex]
Combining these:
[tex]\[ E(U) = \frac{3}{2} \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{3}{2} \times \frac{1}{4} = \frac{3}{8} \][/tex]
Thus, the expected value [tex]\( E(U) \)[/tex] is [tex]\( \frac{3}{8} \)[/tex].
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