The Probability Integral Transform states that if we evaluate the inverse [[Cumulative Density Function|CDF]] of any continuous distribution $X$, denoted as $F^{-1}_X$, and evaluate it at points that are generated by a standard [[Continuous Uniform Distribution]] $Y \sim \mathcal U(0,1)$,we will get data points, following a distribution of $X$. **Key Steps:** 1. *Forward transform:* Sampling from $X$ and applying its CDF $F_X$​ transforms the data into a standard Uniform distribution $Y\sim \mathcal U(0,1)$. $ Y \sim F_X(X) $ 2. *Reverse transform:* Sampling from a Uniform distribution $\mathcal U (0,1)$ and applying the inverse CDF $F_X^{-1}$ generates samples that follow the original distribution $X$. $ X \sim F_X^{-1}(Y) $ **Proof:** To prove that the reverse transform $F^{-1}_X(Y)$ generates samples from $X$, consider the following: $ \begin{align} F_Y(y) &= \mathbf P(Y \le y) \tag{1}\\[2pt] &= \mathbf P(F_X(X) \le y) \tag{2}\\[2pt] &= \mathbf P(X \le F^{-1}_X(y)) \tag{3}\\[2pt] &= F_X(F^{-1}_X(y)) \tag{4}\\[2pt] &=y \tag{5} \end{align} $ where: - (2) Substitute $F_X(X)$ for $Y$. - (3) Use the monotonic property of $F_X$, where $F_X(X) \le y \iff X \le F_X^{-1}(y)$. - (4) Recognize that this can be expressed as a CDF of $X$. - Since $F_X^{-1}$ is the inverse of $F_X$, they offset each other, leaving $y$ as the remainder. By showing that $F_Y(y)=y$ we have proven that it is a standard Uniform distribution. ![[probability-integral-transform.png|center|400]]