## Related questions with answers

Let U and V be independent random variables with means $\mu$ and variances $\sigma^{2}$. Let $Z=\alpha U+V \sqrt{1-\alpha^{2}}$. Find E(Z) and $\rho_{U Z}$.

Solution

VerifiedWe are given that $U$ and $V$ are independent random variables such that $E(U)=E(V)=\mu$ and $\text{Var}(U) = \text{Var}(V) = \sigma^2.$ A random variable $Z$ is defined as

$Z=\alpha U + V\sqrt{1-\alpha^2},$

for some $\alpha \in \mathbb [-1,1],$ and we need to find the expected value of $Z$ and the correlation coefficient of $U$ and $Z.$

First, remember the linearity property of the expectation, and use it to find the expected value of $Z:$

$\begin{align*} E(Z) & = E(\alpha U + V\sqrt{1-\alpha^2}) = \alpha \cdot E(U) + \sqrt{1-\alpha^2}\cdot E(V) \\\\&= \alpha\cdot\mu + \mu\cdot\sqrt{1-\alpha^2} = \boxed{\mu\cdot(\alpha + \sqrt{1-\alpha^2})}\, . \end{align*}$

Next, remember that the correlation coefficient of $U$ and $Z$ is defined as

$\rho(U, Z) = \frac{\text{Cov}(U, Z)}{\sqrt{\text{Var}(U)\cdot \text{Var}(Z)}}\, .$

Let's first find the covariance of $U$ and $Z.$ Due to the bilinearity property of the covariance, we have that

$\begin{align*} \text{Cov}(U, Z) & = \text{Cov}(U, \, \alpha U + V\sqrt{1-\alpha^2}) \\\\ & = \alpha\cdot\underbrace{\text{Cov}(U, U)}_{=\text{Var}(U)=\sigma^2} + \sqrt{1-\alpha^2}\cdot\underbrace{\text{Cov}(U, V)}_{=0} \\ & = \alpha\cdot \sigma^2. \end{align*}$

Here we were able to conclude that $\text{Cov}(U, V) = 0$ because $U$ and $V$ are independent.

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