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Let $X$ be a random variable with c.d.f. $F$ and quantile function $F^{-1}$. Assume the following three conditions:

(i) $F^{-1}(p) = c$ for all $p$ in the interval $(p_0, p_1)$,

(ii) either $p_0 = 0$ or $F^{-1} (p_0) < c$, and

(iii) either $p_1 = 1$ or $F^{-1}(p) > c$ for $p > p_1$.

Prove that $Pr(X=c) = p_1 - p_0$.

No idea how to start proving this condition. Any help will be greatly appreciated. Thanks in advance!

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1 Answer 1

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Here is my interpretation of the question.

Since $F$ is right-continuous, $F^{-1}(p)=c$ for all $p\in(p_{0},p_{1}]$ implies that $F$ has a jump at the point $X=c$. It helps to draw the picture. Note, for any $p>p_{1}$, $F^{-1}(p)>c$, by iii) so we are justified at considering $p\in(p_{0},p_{1}]$.

Also note that the value of $c$ can be viewed as $p_{1}$-quantile: $c=F^{-1}(p_{1})=\inf\{x\in\mathbb{R}:F(x)\geq p_{1}\}.$

So $F(c)=P(X\leq c)=p_{1}.$ Then think about the probability that $P(X<c)$, i.e., probability that $X$ is strictly less than $c.$ (hint: look at condition ii)

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