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I am having trouble understanding this notion. How is a distribution function of a random variable a random variable, because we already have the distribution function determined. Also, more importantly, if $X$ is a random variable, how would we find $D(X)$ where $D(a)$ is $Pr (X\leq a)$. I am unable to wrap my mind around this.

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"How is a distribution function of a random variable a random variable" Basically... it isn't. Where did you get that idea? –  leonbloy Apr 12 '12 at 1:42
    
@RahulNarain : I think your assertion that $D(X)= \Pr(X\le X)$ is simply missing the point. –  Michael Hardy Apr 12 '12 at 1:53

1 Answer 1

One case in which the distribution of a random variable is itself random is this: First choose a number $p$ between $0$ and $1$ randomly---say with a uniform distribution. Then toss a biased coin with probability $p$ of heads. The conditional probability distribution of the outcome of the coin toss is itself chosen randomly.

Your other question is quite a different thing.

Now suppose $X$ has a standard normal distribution. It is conventional to denote the cumulative probability distribution function for this distribution by the capital Greek letter $\Phi$ (and the density by the lower-case $\varphi$). So we have $$ \Pr(X \le x) = \Phi(x), $$ where lower-case $x$ can be any real number at all. If one chooses $X$ randomly from that distribution and then evaluates $\Phi(X)$, then $\Phi(X)$ is itself a random variable. Notice that $\Phi(X)$ is always between $0$ and $1$. And for (lower-case) $x$ between $0$ and $1$ we have (since $\Phi$ is a strictly increasing function) $$ \Pr(\Phi(X) \le x) = \Pr (X \le \Phi^{-1}(x)) = \Phi\Big( \Phi^{-1}(x)\Big) = x. $$ Therefore $\Phi(X)$ is actually uniformly distributed between $0$ and $1$.

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