# deriving $\operatorname{var}(X)=\mathbb{E}(L)(\operatorname{var}(D))+\mathbb{E}(D)^2(\operatorname{var}(L))$

Assume $X$ is continuous random variable representing demand during lead time with density function $f(x)$ and mean, variance, standard deviation $\mathbb{E}(X),\operatorname{var}(X),\sigma_X$ respectively. We assume if $D$ is normally distributed, then so is $X$. Let the lead time denoted $L$ be a random variable as well with mean, variance, and st. dev defined similarly as above. If the length of the lead time is independent of demand per unit time during the lead time, then:

$\operatorname{var}(X)=\mathbb{E}(L)(\operatorname{var}(D))+\mathbb{E}(D)^2(\operatorname{var}(L))$

I would like to know why this is true. In my book it is just stated as a result without explaining why (unless I'm just missing it?)

In case it is important, the context is for ordering supplies in an inventory ordering model where demand is uncertain.

Thanks.

Just some intuition would be appreciated.

HINT:

$X=LD$, where $L$ and $D$ are independent rvs. Since $L$ and $D$ are independent, $E(X)=E(L)E(D)$ because $COV(L,D) = 0$ since they are independent (note the converse is not true).

Now, $Var(X)=E(X^2)-(E(X))^2 = E(L^2D^2) - E(L)^2E(D)^2$. Since $L,D$ are independent, so are their squares. Hence:

$$E(D^2L^2)=E(D^2)E(L^2) = [Var(D)+E(D)^2][Var(L)+E(L)^2] = Var(D)Var(L)+Var(D)E(L^2)+Var(L)E(D)^2+E(D)^2E(L)^2$$

Plugging this into the $Var(X)$ equation, we get:

$$Var(X) = Var(D)Var(L)+Var(D)E(L)^2+Var(L)E(D)^2$$

This is just the variance of the product of two independent random variables.

However, your expression is not quite so symmetric....you'll need to take it from here..