Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.