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I'm just wondering how to prove that $P ( X < F^{-1} (y)) \leq y $ where $F^{-1} (y) = \inf \{x: F(x) \geq y \}$ and $F$ is CDF of random variable X.

I'm sure this is pretty simple, but I can't figure this thing out.

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    Hint: Begin by sketching a graph of an $F(x)$, a _strictly increasing continuous_ function. Then the inverse function $F^{-1}(y)$ is also one-to-one. Is the result obvious in this case? Now modify your graph so that $F(x)$ has constant value on some interval $[a,b)$. Now the inverse function is not one-to-one, and one possible definition for $F^{-1}(y)$ is the one that has been given to you. Does the result hold for this definition?2012-10-22

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Write $P(X As $F^{-1}(y)-n^{-1}, we have $F(F^{-1}(y)-n^{-1}), which gives the inequality (it's large as we took a limit).

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    I know, but here: http://math.stackexchange.com/questions/67287/quantile-function-properties first answer, Fact 3 with proof.2012-10-22