Density of $W=\frac{X+Y}{2}$

Question

Let $X$ a random variable with uniform distribution in $(1,a)$. Knowing that $\mathbb{E}(X)=6Var(X)$,

1) find the value of $a$.

2) find the distribution of $Y=-log(\frac{X-1}{a-1})$.

3) if $Z$ is another variable with the same distribution of $X$ and indipendent of $X$, find the law of $W=\frac{X+Y}{2}$.

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1) $a=3$ because the second solution ($a=0$) is not acceptable for the positivity of density.

2) $f_Y(y)=\mathbb{P}(X\geq 1+(a-1)e^{-y})=\int_{1+(a-1)e^{-y}}^{a}\frac{1}{a-1}dx=1-e^{-y}\Rightarrow Y\sim Exp(1)$.

3) The problem is here. I did $F_W(w)=\mathbb{P}(W\leq w)=\mathbb{P}(X\leq 2w-Z)=\int_{1}^{a}[\int_{1}^{2w-z}\frac{1}{(a-1)^2}dx]dz$ but the result is not correct. I don't understand where I wrong. Could you please help me?

Thanks in advance!

Answer 1

The domain of integration is not correct.

$z$ doesn't always run from $1$ to $a$. It runs from $1$ to $\min(a, 2w-1)$.

Also, we only integrate $x$ from $1$ to $\min(a, 2w-z)$.

The pictures illustrate the cases to consider: