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Let U be a uniform random variable on [0,1]. Let X=$U^2$. Find the distribution function of X.

So i'm confused but I think this is the way to go. F(x)=${\omega \in \Omega; X(\omega) \leq x}. x\in \mathbb{R}$

Since X=$U^2$ the the distribution function is

F(x)= \begin{cases} 0, & \text{if } x < 0, \\ 0, & \text{if } 0 \leq x < 1 \\ 1, & \text{if } x \geq 1 \end{cases}

Any help?

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closed as off-topic by Did, астон вілла олоф мэллбэрг, Daniel W. Farlow, Claude Leibovici, user91500 Feb 7 '17 at 12:40

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What you are saying, is that $U^2 = 1$ always. I'll lay out two approaches.

USING THE DISTRIBUTION FUNCTION

Recall that the CDF for $U$ is:

$F_U(u) = \begin{cases} 0 & u < 0 \\ u & 0 \leq u \leq 1 \\ 1 & u > 1 \end{cases}$

So to find the CDF for $X$, we could do something like:

$F_X(x) = P(X \leq x) = P(U^2 \leq x) = P(U \leq +\sqrt{x}) = F_U(\sqrt{x})$

Hence,

$F_X(x) = \begin{cases} 0 & x < 0 \\ \sqrt{x} & 0 \leq x \leq 1 \\ 1 & x > 1 \end{cases}$

Now, if you wanted to find the Probability Density Function, you need only differentiate the CDF.

$f_X(x) = \frac{1}{2\sqrt{x}} \ , \ x \in [0,1]$

USING TRANSFORMATION OF VARIABLES (PDF's)

We can use the density function of $U$, and the fact that the function $g(u) = u^2$ is monotonically increasing on the interal [0,1] to use the transformation of variables.

$X = U^2 \Rightarrow U = + \sqrt{X}$

The Jacobian in 1D is just the derivative.

$|J| = \frac{1}{2\sqrt{x}}$

Applying the change of variables formula:

$f_X(x) = |J|\cdot f_U(g^{-1}(x)) = \frac{1}{2\sqrt{x}}\cdot 1 = \frac{1}{2\sqrt{x}} $

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    $\begingroup$ so we have to solve for u, that is how you got $\sqrt{2}$ (on Distribution function). I get it now, thank you $\endgroup$ – Killercamin Feb 6 '17 at 20:06
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You are looking for the probability that $X\le x$ for some value of $x$, i.e. the probability that $U^2\le x$, which is also the probability that $U\le\sqrt x$, which is simply $\sqrt x$ as $U$ is uniform (in $[0,1]$).


More generally the cdf of $g(U)$ will be given by $g^{-1}(x)$, provided $g$ is invertible.

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If $U \sim \operatorname{Uniform}(0,1)$, then $$\Pr[U \le u] = \begin{cases} 0, & u \le 0 \\ u, & 0 < u \le 1 \\ 1, & 1 < u. \end{cases}.$$ Then if $X = U^2$, it is clear that the support of $X$ is on $[0,1]$ also, hence for $0 \le x \le 1$, how would we find $$\Pr[X \le x] = \Pr[U^2 \le x] = \Pr[U \le \sqrt{x}]$$ using the cumulative distribution function we wrote for $U$ above?

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