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I seem to be getting it equal to the $X$, which can't be right

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2 Answers 2

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If $X$ is uniform and $0 \leq t \leq 1$ then $$\Pr[X^2 \leq t] = \Pr[X \leq \sqrt{t}] = \sqrt{t}.$$ So $X^2$ is not distributed uniformly.

Intuitively, squaring a number between $0$ and $1$ reduces it, so $X^2$ typically takes smaller values than a uniform $[0,1]$ random variable.

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Whoops. Seems like I can't square root numbers correctly. I imagine taking t^0.5 not equal to +/- t might aid in completing this question. Cheers. –  Freeman Jun 9 '11 at 13:17

First note that $X^2$ is supported on $[0,1]$. For any $x \in [0,1]$, $$ {\rm P}(X^2 \le x) = {\rm P}(X \le \sqrt x ) = \sqrt x . $$ Hence, $X^2$ is characterized by a distribution function $F$ given by $F(x)=0$ if $x < 0$, $F(x)=\sqrt{x}$ if $0 \leq x \leq 1$, and $F(x)=1$ if $x > 1$.

EDIT: The distribution of $X^2$ is the Beta distribution with parameters $a=1/2$ and $b=1$.

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