Expectation of Absolute Difference

Solution & Explanation

To determine the expected value of the absolute difference $|x-y|$ where $x\sim U(0,1)$ and $y\sim U(0,2)$, we need to consider the joint distribution of $x$ and $y$ and evaluate the expectation over the defined support.

Step-by-Step Solution

1. Define the Probability Density Functions (PDFs):

• $x\sim U(0,1)$ implies the PDF of $x$ is $f(x) = 1$ for $0 \leq x \leq 1$.
• $y\sim U(0,2)$ implies the PDF of $y$ is $f(y) = \frac{1}{2}$ for $0 \leq y \leq 2$.

2. Joint PDF:

• Since $x$ and $y$ are independent, the joint PDF is: $f(x,y) = f(x) \cdot f(y) = 1 \cdot \frac{1}{2} = \frac{1}{2}$

3. Expected Value of $|x-y|$:

• The expected value can be calculated by integrating over the entire region where $x$ and $y$ are defined: $E[|x-y|] = \int_{0}^{2} \int_{0}^{1} |x-y| \cdot \frac{1}{2} \ dx \ dy$

4. Split the Integral into Two Cases:

• Case 1: $x > y$ $E[|x-y|] = \int_{0}^{2} \int_{0}^{y} (y-x) \cdot \frac{1}{2} \ dx \ dy$
• Case 2: $x < y$ $E[|x-y|] = \int_{0}^{2} \int_{y}^{1} (x-y) \cdot \frac{1}{2} \ dx \ dy$

5. Evaluate the Integrals:

• For Case 1: $\int_{0}^{2} \int_{0}^{y} (y-x) \cdot \frac{1}{2} \ dx \ dy = \int_{0}^{2} \left[\frac{y^2}{2} - \frac{yx}{2} \right]_{0}^{y} \ dy = \int_{0}^{2} \frac{y^2}{2} \ dy = \frac{8}{12}$
• For Case 2: $\int_{0}^{2} \int_{y}^{1} (x-y) \cdot \frac{1}{2} \ dx \ dy = \int_{0}^{2} \left[\frac{x^2}{2} - \frac{xy}{2} \right]_{y}^{1} \ dy = \int_{0}^{2} \left(\frac{1}{2} - \frac{y}{2} - \frac{y^2}{2} \right) dy = \frac{4}{12}$

6. Combine the Results:

• Add the results from both cases: $E[|x-y|] = \frac{8}{12} + \frac{4}{12} = \frac{12}{12} = \frac{2}{3}$

Conclusion

The expected value of the absolute difference $|x-y|$ is $\frac{2}{3}$. This result is consistent with both analytical methods and simulation approaches, confirming the accuracy of the calculation.