I have no idea how to do this since I don't have the specific function, and prof says that it's the point in p.d.f. as it is $f(x)$.
This is the problem:
The random variable $x$ is distributed according to the p.d.f. $f(x)$. For two intervals $[F_1, F_2]$ and $[F_3,F_4]$, $0≤F_i≤1$, compute the ratio of probabilities for the cumulative density function $F(x)$ of the function $f(x)$ being in one or the other interval, i.e: $$R=\frac{\text{Prob}(F_1≤F(x)≤F_2)}{\text{Prob}(F_3≤F(x)≤F_4)}$$
Please Help!!!
Thx!
Let $X$ be a random variable with pdf $f$ and CDF $F$. Solving this problem requires you to work out the distribution of $Y$ if $Y \equiv F(X)$. For simplicity I assume that $F$ is strictly increasing, so that the quantile function $F^{-1}$ is strictly increasing. Then we have: \begin{eqnarray*} P(Y \leq y) &= P\left[ F(X) \leq y \right] = P\left[ F^{-1}\big(F(X) \big)\leq F^{-1}(y) \right]\\ &= P\left[ X \leq F^{-1}(y)\right]\\ &= F\big( F^{-1}(y))= y \end{eqnarray*} Hence, we see that $Y$ is a Uniform$(0,1)$ random variable. The preceding argument is often called the "Probability Integral Transform." From here we obtain $R = (F_2 - F_1)/ (F_4 - F_3)$.