MHB Probability Homework Question 2

[pinky14]

Suppose that X, Y are uncorrelated random variables which are each measurements of some unknown quantity $\mu$. Both random variables have $\mu_{X} = \mu_{Y} = \mu$, but $\sigma^2_{X} > \sigma^2_{Y}$. Determine the value of $\alpha$ in [0, 1] which will minimize the variance of the random variable W = $\alpha X +(1 - \alpha)$Y. Note that E(W) = $\mu$ for any $\alpha$ so the minimal variance α gives the “best” linear combination of X, Y to use in estimating $\mu$.

 

[Euge]

Hi pinky14,

In the future, please show your thoughts or what you've tried.

Show that $\sigma_W^2 = \alpha^2 \sigma_X^2 + (1 - \alpha)^2\sigma_Y^2 + 2\alpha(1 - \alpha)\operatorname{Cov}(X,Y)$. Since $X$ and $Y$ are uncorrelated, what can you say about $\operatorname{Cov}(X,Y)$?

 

[pinky14]

Hi pinky14,

In the future, please show your thoughts or what you've tried.

Show that $\sigma_W^2 = \alpha^2 \sigma_X^2 + (1 - \alpha)^2\sigma_Y^2 + 2\alpha(1 - \alpha)\operatorname{Cov}(X,Y)$. Since $X$ and $Y$ are uncorrelated, what can you say about $\operatorname{Cov}(X,Y)$?

So for 2 random variables to be "uncorrelated," their covariance has to be 0? (E(XY)-E(X)E(Y))? I wasn't really sure how to start with this problem. The problem says their means are equal but that the variance of X is larger than Y but I am not sure what to do with the equation for W that they give. When they say to minimize, do we take the derivative? I am pretty lost in my probability class.

 

[Euge]

So for 2 random variables to be "uncorrelated," their covariance has to be 0?

That's correct. So then $\sigma_W^2 = \alpha^2 \sigma_X^2 + (1 - \alpha)^2 \sigma_Y^2$. You will need to find $\alpha\in [0,1]$ that minimizes the quadratic function $f(\alpha) := \alpha^2 \sigma_X^2 + (1 - \alpha)^2 \sigma_Y^2$.