I try to prove:

Let $X$ be a discrete random variable with cdf $F$, $F_{-}(x):=P(X<x)$, $W \sim U[0,1]$ a random variable and $X, W$ independent.

Then $$V:=WF(X)+(1-W)F_{-}(X) \sim U[0,1].$$

My ideas: $$F_V(v|X=x)=P(V\leq v|X=x)=\frac{P(\lbrace V\leq v\rbrace \cap \lbrace X=x \rbrace)}{P(X=x)}$$ and since $$V\leq v \Leftrightarrow WF(X)+(1-W)F_{-}(X) \leq v \Leftrightarrow W \leq \frac{v-F_{-}(X)}{F(X)-F_{-}(X)}$$ we get $$F_V(v|X=x)=\frac{P(\lbrace W \leq \frac{v-F_{-}(X)}{F(X)-F_{-}(X)}\rbrace \cap \lbrace X=x \rbrace)}{P(X=x)}.$$ Because $X,W$ independent: $$F_V(v|X=x)=\frac{P(W \leq \frac{v-F_{-}(X)}{F(X)-F_{-}(X)})P( X=x )}{P(X=x)}.$$ And since $W\sim U[0,1]$ we finally get $$F_V(v|X=x)=\frac{v-F_{-}(X)}{F(X)-F_{-}(X)}=\frac{v-F_{-}(X)}{P(X=x)}.$$

We can conclude: $$F_V(v)=P(V\leq v)=\sum\limits_x P(V\leq v |X=x)P(X=x)=\sum\limits_x v-F_{-}(x)$$

So my question now is: What's wrong about this? And how can I deduct that $V \sim U[0,1]$?


1 Answer 1


Your calculation of $P(V\le v|X=x)$ needs to be corrected slightly. It should be $\min((v-F_-(x))^+/f(x),1)$. (Here, $c^+=\max(c,0)$. In effect, the conditional distribution of $V$, given that $X=x$, is uniform on the interval $[F_-(x),F(x)]$, and the cdf of a random variable that is uniform on $[a,b]$ is given by $x\mapsto \min((x-a)^+/(b-a),1)$. With this change the sum works out correctly.

  • $\begingroup$ So I get $F_V(v)=\sum\limits_x \min{(\frac{(v-F_{-}(x))^+}{f(x)},1)}f(x)=\sum\limits_x \min{(v-F_{-}(x))^+,f(x))}$, but how does the sum work out now? I know now that the result should be $\min(z^+,1)$ but I really don't see how I can simplify the sum. $\endgroup$
    – user154085
    Nov 3, 2015 at 8:56
  • $\begingroup$ I found it hekpful to draw a picture of the cdf $F$. $\endgroup$ Nov 4, 2015 at 4:49

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