For any random variable $X$ taking values in $[0,1]$ with distribution $\mu_X$, and $f$ defined by $f(t) = \int_{-\infty}^t d\mu_X$, I'm stuck trying to show that $$P(f(X) \leq a) = a$$ for all $a \in [0,1]$. I don't really have any work to show. I've tried rewriting things using integrals, but it hasn't gotten me anywhere. Even just a hint would be helpful.
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I don't understand the question. What is $f(X)$? Is it $P(X≤X)$? that would seem to be $1$. Is $f(X)$ somehow meant to be a random variable? – lulu Jun 16 '17 at 13:05
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If $a=39$ your expression would appear to give a probability equal to $39$. Just for example. – lulu Jun 16 '17 at 13:06
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@lulu Sorry, edited – user435571 Jun 16 '17 at 13:07
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Much better, but I still don't get it. If $a=0$ your expression would appear to suggest that the probability that $X≤0$ is $0$, but that need not be true. – lulu Jun 16 '17 at 13:13
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@lulu Sorry, edited again. – user435571 Jun 16 '17 at 13:15
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Ah, now I believe it. – lulu Jun 16 '17 at 13:18
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2Possible duplicate of Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X – NCh Jun 18 '17 at 09:07
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The function $f$ is the cumulative distribution function (CDF) of $X$, defined by $t \mapsto \mathbb{P}(X\leq t)$. By definition, $f$ is increasing. Moreover, if $f$ is continuous and strictly increasing, then \begin{aligned} \mathbb{P}(f(X)\leq a ) & = \mathbb{P}(X\leq f^{-1}(a) ) \\ & = f(f^{-1}(a))\\ & = a \end{aligned} for $a$ is in $[0,1]$. If there is a single subdomain $]x_\mathit{inf},x_\mathit{sup}[$ of $\mathbb{R}$ on which $f$ is constant, then \begin{aligned} \mathbb{P}(f(X)\leq a ) &= {\left\lbrace \begin{aligned} &\mathbb{P}(X\leq f^{-1}(a) ) && \text{if} \quad a < f (x_\mathit{sup}) \\ &\mathbb{P}(X\leq x_\mathit{sup} ) && \text{if} \quad a = f (x_\mathit{sup}) \\ &\mathbb{P}(X\leq f^{-1}(a) ) && \text{if} \quad f (x_\mathit{sup}) < a \, . \end{aligned}\right.} \\ &= a \, . \end{aligned} Therefore, the property still holds. Now let us assume that $f$ has a single discontinuity at the abscissa $x_0$. If the jump in $f$ is of height $f (x_0^+) - f (x_0^-)$, then \begin{aligned} \mathbb{P}(f(X)\leq a ) &= {\left\lbrace \begin{aligned} &\mathbb{P}(X\leq f^{-1}(a) ) && \text{if} \quad a < f (x_0^-) \\ &\mathbb{P}(X\leq x_0 ) && \text{if} \quad f (x_0^-) \leq a < f (x_0^+) \\ &\mathbb{P}(X\leq f^{-1}(a) ) && \text{if} \quad f (x_0^+) \leq a \end{aligned}\right.} \\ &= {\left\lbrace \begin{aligned} &f (x_0^-) && \text{if} \quad f (x_0^-) \leq a < f (x_0^+) \\ &a && \text{otherwise} \, . \end{aligned}\right.} \end{aligned} In this case, the property is no longer true.
To summarize, if the CDF $f$ is continuous, i.e. X is a continuous random variable, then $f(X)$ has a uniform distribution.
EditPiAf
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If $f$ is strictly increasing, then Harry49's answer works just fine. If not, this is no longer true. Take the simple example where $P(X=0)=P(X=1)=\frac12$. Then $f(x)=\frac12$ for $0\le x<1$ and $f(1)=1$. Take $a=\frac14$. Then $P(f(X)\le a)=0\neq a$.
Jason
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