Problem Definition
I want to find the variance of the RV $Z=(B-A)/2$, where $A$, $B$ are RVs with joint distribution
$$f_{AB}(a,b)=\begin{cases} \frac{p(p-1)}{(2\delta)^p}(b-a)^{p-2} & \text{if } (a,b)\in T_\delta \\ 0 & \text{elsewhere}\end{cases}$$
defined over the support $T_\delta=\{(a,b)\in \mathbb{R}^2: a<b, |a|<\delta, |b|<\delta\}$. The correct answer is $$\text{Var}[Z]=\frac{2}{(p+1)(p+2)}\delta^2$$
Proposed Solution
My strategy consist in the following 3 steps:
- To find the CDF of $Z$ with the formula $$\begin{align*}F_Z (t) =\mathbb{P}({Z\leq t}) &=\int_{D(t)} f_{A,B}(a,b)\text{ d}a\text {d}b \\ &=\frac{p(p-1)}{(2\delta)^p} \int_{D(t) \cap T_\delta} (b-a)^{p-2}\text{ d}a\text {d}b \end{align*}$$ where $D(t)=\{(a,b)\in\mathbb{R}^2:Z\leq t\}=\{(a,b)\in\mathbb{R}^2:B\leq A+2t\}$.
Then, to find by differentation the PDF of $Z$, i.e. $$f_Z (t) =\frac{\text{d}F_Z}{\text{d}t}(t)$$ - To calculate the expected values of the RVs $Z$ and $Z^2$ by appling the formulae $$\mathbb{E}[Z]=\int_{\mathbb{R}} t f_Z (t)\text{ d}t \qquad \mathbb{E}[Z^2]=\int_{\mathbb{R}} t^2 f_Z (t)\text{ d}t$$
- To calculate the variance through the formula $$\text{Var}[Z]=\mathbb{E}[Z^2]-(\mathbb{E}[Z])^2$$
First Step Quite trivially it turns out that $F_Z (t)=0$ for $t<0$ and $F_Z (t)=1$ for $t\geq \delta$. For $0\leq t<\delta$ the CDF get the values $$\begin{align*}F_Z (t) &=\frac{p(p-1)}{(2\delta)^p}\left[\int_{-\delta}^{-\delta+2t}\int_{-\delta}^{+b}(b-a)^{p-2}\text d{a}\text{d}b+\int_{-\delta+2t}^{+\delta}\int_{b-2t}^{+b}(b-a)^{p-2}\text d{a}\text{d}b\right] \\ &=\frac{p}{(2\delta)^p}\left[\int_{-\delta}^{-\delta+2t}(b+\delta)^{p-1}\text{d}b+\int_{-\delta+2t}^{+\delta}(2t)^{p-1}\text{d}b\right] \\ &=\frac{p}{(2\delta)^p}\left[\frac{(2t)^p}{p}+(2t)^{p-1}(2\delta-2t)\right] \\ &=\frac{1-p}{\delta^p}t^p+\frac{p}{\delta^{p-1}}t^{p-1} \end{align*}$$ thus the PDF is $$f_Z(t)=\begin{cases} \frac{1-p}{\delta^p}pt^{p-1}+\frac{p}{\delta^{p-1}}(p-1)t^{p-2} & \text{if } 0\leq t < \delta \\ 0 & \text{elsewhere}\end{cases}$$
Second Step
$$\begin{align*} \mathbb{E}[Z] &= \int_{0}^\delta \frac{1-p}{\delta^p}pt^{p}+\frac{p}{\delta^{p-1}}(p-1)t^{p-1} \text{ d}t \\
& = \frac{p(1-p)}{\delta^p}\frac{\delta^{p+1}}{p+1}+\frac{p(p-1)}{\delta^{p-1}}\frac{\delta^p}{p} \\
&=\frac{p-1}{p+1}\delta
\end{align*}$$
$$\begin{align*} \mathbb{E}[Z^2] &= \int_{0}^\delta \frac{1-p}{\delta^p}pt^{p+1}+\frac{p}{\delta^{p-1}}(p-1)t^{p} \text{ d}t \\
& = \frac{p(1-p)}{\delta^p}\frac{\delta^{p+2}}{p+2}+\frac{p(p-1)}{\delta^{p-1}}\frac{\delta^{p+1}}{p+1} \\
&=\frac{p(p-1)}{(p+1)(p+2)}\delta^2
\end{align*}$$
Third Step $$\begin{align*} \text{Var}[Z] &=\frac{p(p-1)}{(p+1)(p+2)}\delta^2 -\left(\frac{p-1}{p+1}\delta\right)^2 \\ &=\frac{2(p-1)}{(p+1)^2(p+2)}\delta^2 \end{align*}$$ the result is incorrect because the sign of the numeric term $-1$ at the numerator. I've checked the calculation, but I can't find the error.