Let $f\left( x \right) = {{\left( {1 + \alpha x} \right)} \over 2}$
for $ – 1 \le x \le 1$, and $0$ otherwise. We further assume that
$\left| \alpha \right| < 1$. Show that (i) $f$ is a density, (ii)
find the cumulative distribution $F$ and (iii) the quantile function $Q$.
My attempt:
Items (i) and (ii) were straightforward.
(i) I showed that $$\int\limits_{ – 1}^1 {{{\left( {1 + \alpha x} \right)} \over 2}} dx = {1 \over 2}\left( 2 \right) + {1 \over 4}\left( {1 – 1} \right) = 1$$
so $f$ is a density indeed.
(ii) I found $F$ by doing $$\int\limits_{ – 1}^x {{{\left( {1 + \alpha t} \right)} \over 2}dt} = F\left( x \right) = {1 \over 4}\left( {\alpha {x^2} + 2x + 2 – \alpha } \right)$$
Until here, both answers match the final answers, which were provided.
(iii) Regarding the quantile function, I can't finish my attempt. Knowing that the quantile function is the inverse of the CDF, I wrote ${1 \over 4}\left( {\alpha {x^2} + 2x + 2 – \alpha } \right) = y$ and tried soving for $x$ in order to find a $F(p)=x$.
$$\eqalign{
& \alpha {x^2} + 2x + \left( {2 – \alpha – 4y} \right) = 0 \cr
& \Delta = {b^2} – 4ac = 4 – 8\alpha + 4{\alpha ^2} + 16\alpha y \cr} $$
I knew this delta would be problematic and I couldn't go past this point:
$$\eqalign{
& x = {{ – b \pm \sqrt \Delta } \over {2a}} \cr
& {x_1} = {{ – 2 + \sqrt {4 – 8\alpha + 4{\alpha ^2} + 16\alpha y} } \over {2\alpha }} \cr
& {x_2} = {{ – 2 – \sqrt {4 – 8\alpha + 4{\alpha ^2} + 16\alpha y} } \over {2\alpha }} \cr} $$
I don't know if there's anyway I can "ignore" one of the $x's$ based on given conditions, or if this is not the way to go at all. The final answer is supposed to be $Q\left( \beta \right) = 2\beta – 1$ if $\alpha = 0$ and $Q\left( \beta \right) = {1 \over \alpha }\left( { – 1 + \sqrt {{\alpha ^2} + 4\alpha \beta – 2\alpha + 1} } \right)$ if $\alpha \ne 0$. (I'm not even sure I got this beta notation).
Best Answer
The point is that one of the roots of the quadratic isn't even in $[-1,1]$. Evidently the root that is in $[-1,1]$ is the one you called $x_1$, indeed their solution agrees with your $x_1$ once you cancel out a factor of $2$.