Why does the variance of =2Θ+3 , where Θ and are standard normal random variables, is equal to 9?

Question

I dont understand why the variance is equal to 9 and actually why the mean is $2\cdot \theta$.

Answer 1

It isn't. $$ \mathbb{E}(2\Theta+3W) = 2\mathbb{E}(\Theta)+3\mathbb{E}(W)=2\cdot 0 +3\cdot 0=0.$$ If $\Theta$ and $W$ are independent, $$\mathbb{V}(2\Theta+3W) = 2^2\mathbb{V}(\Theta)+3^2\mathbb{V}(W)=4\cdot 1 +9\cdot 1=13.$$

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If instead $\Theta$ is constant and $W$ is standard normal, $$ \mathbb{E}(2\Theta+3W) = 2\mathbb{E}(\Theta)+3\mathbb{E}(W)=2\Theta +3\cdot 0=2\Theta$$ and $$\mathbb{V}(2\Theta+3W) = 2^2\mathbb{V}(\Theta)+3^2\mathbb{V}(W)=4\cdot 0 +9\cdot 1=9.$$

Answer 2

Assuming you meant for $\theta$ to be constant (I've decapitalised it because statisticians reserve upper case for non-constant random variables),$$\operatorname{Var}(2\theta+3W)=\operatorname{Var}(3W)=3^2\operatorname{Var}W=9\times1=9.$$