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Suppose $X$ is distributed uniformly on $(−1,1)$. For $i = 0, 1, 2, 3$, let $p_i = P\bigg(X^{2} \in \bigg(\frac{i}{4}, \frac{i+1}{4}\bigg)\bigg)$. For which value of $i$ is $p_i$ the largest?

(A) 3

(B) 1

(C) 0

(D) 2

What I want to ask is since $X$ follows uniform distribution on (−1,1) what will $X^{2}$ follow?

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We have the equivalence between these events: $$ \left \{ X^2 \in \left ( \frac{i}{4} , \frac{i+1}{4} \right ) \right \} \iff \left \{ X \in \left ( \frac{\sqrt{i}}{2} , \frac{\sqrt{i+1}}{2} \right ) \right \} \bigcup \left \{ X \in \left ( -\frac{\sqrt{i+1}}{2} , -\frac{\sqrt{i}}{2} \right ) \right \} $$

The union on the lhs is disjoint and its probability is the sum of both events.

Moreover, since $X$ is uniformly distributed over $[-1,1]$ then

$$ \mathbb P\left(X \in \left ( -\frac{\sqrt{i+1}}{2} , -\frac{\sqrt{i}}{2} \right ) \right) = \mathbb P \left (X \in \left ( \frac{\sqrt{i}}{2} , \frac{\sqrt{i+1}}{2} \right ) \right) $$

because both intervals have the same length.

Thus, \begin{align*} \mathbb P\left( X^2 \in \left ( \frac{i}{4} , \frac{i+1}{4} \right ) \right ) &= 2 \times \mathbb P \left (X \in \left ( \frac{\sqrt{i}}{2} , \frac{\sqrt{i+1}}{2} \right )\right) \\ &= 2 \times \frac{\frac{\sqrt{i+1}}{2} - \frac{\sqrt{i}}{2}}{2} \\ &= \frac{\sqrt{i+1}-\sqrt{i}}{2} \end{align*}

which is maximal when $i= 0$.

We can also determine the cumulative distribution of $X^2$:

Let $x\in (0,1)$,

\begin{align*} \mathbb P(X^2 \leq x) &= \mathbb P(-\sqrt{x} \leq X \leq \sqrt{x} ) \\ &= \frac{\sqrt{x} - (-\sqrt{x})}{2} \\ &= \sqrt{x} \end{align*}

The density of $X^2$ is then $f(x) = \frac{1}{2\sqrt{x}}$ which is $\propto x^{-\frac{1}{2}}$ and is a particular case of the Beta distribution with $\alpha = \frac{1}{2}$ and $\beta = 1$.

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