Question

Suppose X and Y are independent continuous random variables uniformly distributed on the intervals 0 \leq x \leq 2 and 0 \leq y \leq 4, respectively. Compute the variance of \sqrt{2}X-\frac{1}{2}Y. Hint: First find the variance of X and the variance of Y.

Random variables
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asked 2021-05-18
Suppose X and Y are independent continuous random variables uniformly distributed on the intervals \(0 \leq x \leq 2 and 0 \leq y \leq 4\), respectively. Compute the variance of \(\sqrt{2}X-\frac{1}{2}Y\). Hint: First find the variance of X and the variance of Y.

Answers (1)

2021-05-19
Step 1
Given information
X and Y independent continuous random variables follows uniform distribution
\(0 \leq x \leq 2, 0 \leq y \leq 4\)
Step 2
\(Var(X)=\frac{(2-0)^{2}}{12}=\frac{4}{12}=\frac{1}{3}\)
\(Var(Y)=\frac{(4-0)^{2}}{12}=\frac{16}{12}=\frac{4}{3}\)
Variance of \(ax+by=a^{2}var(x)+b^{2}var(y)\)
\(Var(\sqrt{2}X-\frac{1}{2}Y)=(\sqrt{2})^{2}Var(X)+(-\frac{1}{2})^{2}Var(Y)=2\times \frac{1}{3}+\frac{1}{4}\times \frac{4}{3}=1\)
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