# Proof of inequality (probability, generalized inverse function)

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? –  Dilip Sarwate Oct 22 '12 at 12:42

Write $$P(X<F^{-1}(y))=\lim_{n\to+\infty}P(X\leq F^{-1}(y)-n^{-1})=\lim_{n\to +\infty}F(F^{-1}(y)-n^{-1}).$$ As $F^{-1}(y)-n^{-1}<F^{-1}(y)$, we have $F(F^{-1}(y)-n^{-1})<y$, which gives the inequality (it's large as we took a limit).
But $F(F^{-1}(y)) \geq y$ (not = $y$). Or maybe I'm wrong. I need the inequality in my question to prove that $F^{-1}(y)$ is in fact $y$th quantile of distr. F. –  whoiswho Oct 22 '12 at 13:14
Why $F(F^{-1}(y))\geq y$? We want to show $\leq$? –  Davide Giraudo Oct 22 '12 at 13:18