The variance of $X_1$, $X_2$ are 1 and 4, and the correlation coefficient p=-0.3
1)Calculate the variance of $Z_1 = 2X_1+X_2$
2)Calculate the value of a that minimizes the variance of $Z_2 = aX_1+(1-a)*X_2$
I am currently studying probability and I am not quite sure how to solve the problem stated.
1) Since $Var(a*X_1+b*X_2) = a^2*Var(X_1)+b^2*Var(X_2)+a*b*Cov(X_1,X_2)$ and $p(X_1,X_2) = \frac{Cov(X_1,X_2)}{\sqrt{(Var(X_1)*Var(X_2)}}$
I did the following: $Cov(X_1,X_2) = -0.3*\sqrt{1*4} = -0.6$ and further $Var(Z_1) = 2*1+4+\sqrt{2}*2*-0.6 = 4.3...$
2) For the second problem I am not so sure. Should I use the formula for the variance the same way I did in example 1. Then derive it in respective to a and set it to 0 to get the minimum? When I tried that the derivation looked a bit too complicated, so I am not sure if that was the correct approach.
$Var(Z_2) = a*X_1 + (1-a)*X_2 + 2*\sqrt{a}*\sqrt{1-a}*Cov(X1,X2)$
It would be nice to receive some feedback if my calculations are correct or not. If they are wrong, some explanation would help.
You have made a few mistakes when calculating the variance of random variable $Z_{2}\ .$ $$\text{Var}(Z_{2}) = a^2\text{Var}(X_{1})+(1-a)^2\text{Var}(X_{2})+2a(1-a)\text{Cov}(X_{1},X_{2})\ .$$ This is a quadratic function, $$f(a)=a^2\text{Var}(X_{1})+(1-a)^2\text{Var}(X_{2})+2a(1-a)\text{Cov}(X_{1},X_{2})\ , \quad a\in(0,1)\ ,$$ and by completing the squares you can simultaneously find the smallest value of $f$ and for which $a$ it occurs.
Edit: January 22, 2016.
Denoting by $\sigma_1$ and $\sigma_2$ the standard deviations of $X_{1}$ and $X_{2}$ respectively and by $\rho$ the correlation coefficient of $X_{1}$ and $X_{2}$ the variance $\text{Var}(Z_{2})$ can be expressed as follows. $$a^2\sigma_1^2+(1-a)^2\sigma_2^2+2a(1-a)\rho\sigma_1\sigma_2 = \left\{a\sigma_1+(1-a)\rho\sigma_2\right\}^2+(1-a)^2(1-\rho^2)\sigma_2^2\ .$$ This demonstrates that variance of $Z_{2}$ is minimized when parameter $a$ satisfies the condition $$a\sigma_1+(1-a)\rho\sigma_2=0\ ,$$ in which case the variance equals $(1-a)^2(1-\rho^2)\sigma_2^2\ .$
In the special case where $X_{1}$ and $X_{2}$ are linearly dependent (in which case $\rho^2=1$) variance of $Z_{2}$ becomes zero.