Suppose we have uniform numbers from \( [0,1] \). We want to transform uniform random variables from uniform distribution to our target exponential distribution. We know the probability density function of our target distribution: \[ f(x) = e^{-\lambda x}, x \geq 0 \] and \[ f(x) = 0, x < 0 \] We also know the cumulative density function of our target distribution: CDF tells us, what is the probability that you choose a number that is \( x \) or less from the distribution? \[ F(x) = 1 - e^{-\lambda x}, x \geq 0 \] and \[ F(x) = 0, x < 0 \] Mathematically speaking: What is the transformation \( T \) that we can do to \( U \) in order to get the exponential numbers \( X \)? \[ T(U) = X \] From the cdf definition, \[ F(x) = P(X \leq x) \] In other words, what is the probability that the exponential distribution is less than or equal to small \( x \)? We know that big \( X \) is same as \( T(U) \).We can rewrite above expression as: \[ P(T(U) \leq x) \] Taking the inverse of \( T \) on both sides, we get: \[ P(U \leq T^{-1}(x)) \] What is the probability that the uniform distribution is less than or equals to some value \( T^{-1}(x) \)? In other words, what is the probability that uniform distribution is behind the range \( [0, T^{-1}(x)] \)? What is the probability of the area being behind \( T^{-1}(x) \) = \( 0.25 \). It is \( 0.25 \) because of nice characteristic of uniform distribution (it is uniform all across) Hence, \( F(x) = T^{-1}(x) \) and \[ F^{-1}(x) = T(x) \] We found that cumulative density function for exponential distribution is equals to the inverse of \( T \) of \( x \). In conclusion, \( F \) and \( T \) are inverse functions so if we figure out the inverse of the given density function then we are going to have a transformation we need and apply it to the uniform numbers to get exponential distribution back. Solving for \( x \) to get the inverse: \[ y = 1 - e ^{-\lambda x} \] \[ 1 - y = e ^{-\lambda x} \] \[ -\frac{ln(1-y)}{\lambda} = x \] Let us replace \( y \) with \( u \). \[ -\frac{ln(1-u)}{\lambda} = x \] Since uniform distribution is symmetric, we can re-write as \[ X = -\frac{ln(U)}{\lambda} \] Note: We need cdf in order to apply this method