Find the variance of $R$ where $R$ = $Z_1 + \dotsb + Z_d$ and $Z_i = |X_i - Y_i|^2$

Question

So I am trying to find the Variance $R$ where $R$ = $Z_1 + \dotsb + Z_d$ and $Z_i = |X_i - Y_i|^2$

$X$ and $Y$ are d-dimensional points from a d-dimensional unit cube with a uniform distribution: $X,Y \in [0,1]^d$ which we can view this as drawing random variables $X_1, . . . , X_d$ and $Y_1, . . . , Y_d$ independently and uniformly from $[0, 1]$

Assuming that this is correct: \begin{align*} R &= Z_1 + \dotsb+ Z_d\\ &= d \cdot Z \\ R^2 &= d^2 \cdot Z^2\\ E[R^2] &= d^2 \cdot E[Z^2]\\ &=\frac{12d^2}{180}\\ &=\frac{d^2}{15} \end{align*} and with the information from: Expectation and variance of the squared distance between $X$ and $Y$

I was able to get to: \begin{align*} Var(R) &= E[R^2]-(E[R])^2\\ &=\frac{d^2}{15}-\frac{d^2}{36} \end{align*}

Is this even correct? Or did I make a mistake along the way

Answer 1

$R^2 = \sum_{r=1}^d Z_r^2\ +\ $

$ \sum_{r=1}^d \sum_{r\not=s=1}^d Z_rZ_s$

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$Z_r Z_s$ occurs d(d-1)times and $Z_r^2$ occur d times

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$E(R^2)= d\times E(Z^2)\ +\ d(d-1)\times [E(Z)]^2$ $=\frac{d}{15}\ +\frac{d(d-1)}{36}$

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Variance = $ E(R^2) - [E(R)]^2=\frac{d}{15}\ +\ \frac{d(d-1)}{36}\ -\ \frac{d^2}{36} = \frac{7d}{180}$