# How can one manipulate an equation within variance?

I encountered the following equation to find the variance of, given that $$V(\hat\theta_1)=\sigma^2_1$$,and $$V(\hat\theta_2)=\sigma^2_2$$.

I considered that for the variance of $$\mathcal F=\alpha\hat\theta_1+(1-\alpha)\hat\theta_2$$: $$V(\mathcal F)=V(\alpha\hat\theta_1+(1-\alpha)\hat\theta_2)=V(\alpha\hat\theta_1+\hat\theta_2-\alpha\hat\theta_2)=\alpha^2V(\hat\theta_1)+V(\hat\theta_2)+\alpha^2V(\hat\theta_2)=\alpha^2\sigma^2_1+\sigma^2_2+\alpha^2\sigma^2_2$$

The professor considered: $$V(\mathcal F)=\alpha^2\sigma^2_1+(1-\alpha)^2\sigma^2_2\color{red}{\stackrel{\mathtt{expansion}}{=}\alpha^2\sigma^2_1+(1-2\alpha+\alpha^2)\sigma^2_2=\alpha^2\sigma^2_1+\sigma^2_2-2\alpha\sigma^2_2+\alpha^2\sigma^2_2}$$

I do not see how these two statements are equivalent, presuming both ways are correct in the application of the variance properties.

$$V(\mathcal F)=V(\alpha\hat\theta_1+(1-\alpha)\hat\theta_2)=V(\alpha\hat\theta_1+\hat\theta_2-\alpha\hat\theta_2)\color{red} \neq\alpha^2V(\hat\theta_1)+V(\hat\theta_2)+\alpha^2V(\hat\theta_2)$$
Could you see the reason? $$\hat\theta_2$$ and $$\alpha\hat\theta_2$$ are not uncorrelated.
Please see Basic properties of Variance. In particular, check the formula for $$\text{Var} \left(\sum_{i=1}^N a_i X_i \right)$$
• What do you mean by not uncorrelated? I used the property that $V(aX+b)=a^2V(X)$. Commented Feb 11, 2021 at 0:54