I am trying to calculate the variance of $\int_{t_{1}}^{t_{2}} W(s) \,ds$. Assume $t_{2} \geq t_{1}>0$.
My approach so far has been, \begin{align} & \operatorname{Var}\left[\int_{t_1}^{t_2} W(s) \,ds\right] = \operatorname E\left[\left(\int_{t_1}^{t_2} W(s) \,ds\right)^2\right] \\[6pt] = {} &\int_{t_1}^{t_2} \int_{t_1}^{t_2} E[W(s) W(u)] \,du \,ds = \int_{t_1}^{t_2} \int_{t_1}^{t_2} \min(s,u) \,du \,ds \\[6pt] = {} &\int_{t_1}^{t_2} \int_{t_1}^s u \,du \,ds + \int_{t_1}^{t_2} \int_s^{t_2} s \,du \,ds \\[6pt] = {} &\int_{t_1}^{t_2} \frac{s^2}{2} - \frac{t_1^2}{2} \,ds + \int_{t_1}^{t_2} s(t_2-s) \,ds \\[6pt] = {} & \frac{2t^3_1}{3} + \frac{t^3_2}{3} - t^2_1 t_2 \end{align}
This satisfies the various checks like,
1) If we let $t_{1} = 0$ and $t_{2} = t$ then we get $\frac{t^3}{3}.$
2) If we let $t_{1} = t$ and $t_{2} = t$ then we get $0$.
My question is - Is this the correct answer ? If it is, how do I prove the non-negativity of the variance?
Thank you in advance.
As Michael Hardy mentioned in comments the interesting question here is why the expression $$ \displaystyle \frac{2t^3_1}{3} + \frac{t^3_2}{3} - t^2_1 t_2 \geq 0, $$ for all $0\leq t_1\leq t_2$. We can get the answer by the following arguent. First observe that $$ \displaystyle \frac{2t^3_1}{3} + \frac{t^3_2}{3} - t^2_1 t_2 = t_2^3\left( \frac{1}{3}+\frac{2}{3}\left(\frac{t_1}{t_2}\right)^3 -\left(\frac{t_1}{t_2}\right)^2 \right) = t_2^3\left( \frac{1}{3}+\left(\frac{t_1}{t_2}\right)^2 \left[\frac{2}{3}\left(\frac{t_1}{t_2}\right)-1\right] \right). $$ Since $0\leq t_1/t_2\leq 1$, we can reduce the question to prove that $(1/3)+x^2((2/3)x-1)$ is non-negative for all $x\in [0,1]$. Note that we recover the expression in parenthesis by taking $x=t_1/t_2$. The minimum of the above polynomial can be computed explicitly by taking the derivative. One can see that the minimum is attained at x=1 and the minimum is zero. Since $t_2^3$ is non-negative the result follows.