I'm trying to understand a proof which uses the trick that if $W_t\sim U(0, F(t))$ has a uniform distribution where $F$ is cdf for a random variable $X$, then
$$\mathbb E[\log X|X\leq t]=\mathbb E[\log(F^{-1}(W_t))]$$
I believe $F^{-1}(W_t)\sim F$ in $[0,F(t)]$, and $$\mathbb E[\log X|X\leq t]=\frac{\mathbb E[\log X *1_{X\leq t}]}{P(X\leq t)}=\frac{1}{F(t)}\mathbb E[\log X*1_{X\leq t}]$$ but can't combine them together. Thanks in advance for any help to make it clearer.
We have that $$ \log(F^{-1}(W_t))\sim\log X\mid X\le t $$ for $t>0$. For $x\le\log t$, \begin{align*} P(\log(F^{-1}(W_t))\le x) &=P(F^{-1}(W_t)\le\exp(x))\\ &=P(W_t\le F(\exp(x)))\\ &=\frac{F(\exp(x))}{F(t)}\\ &=\frac{P(X\le\exp(x))}{P(X\le t)}\\ &=\frac{P(\log X\le x)}{P(X\le t)}\\ &=\frac{P(\log X\le x, X\le t)}{P(X\le t)}\\ &=P(\log X\le x\mid X\le t). \end{align*} Since $\log(F^{-1}(W_t))$ has the same distribution as $\log X\mid X\le t$, their expected values are equal.
I hope this helps.