Exponential Distribution

• pdf
  
  $$f(x) = \left\{ \begin{array}{ll} \lambda e^{-\lambda x}, & x \ge 0 \\ 0, & x < 0 \end{array} \right.$$
  
  

• cdf
  
  $$F(x) = \int_{-\infty}^x f(t) dt = \left\{ \begin{array}{ll}... ...e^{-\lambda x}, & x \ge 0 \\ 0, & x < 0 \end{array} \right.$$
  
  

• The idea is to solve $y = 1 - e^{-\lambda x}$ for x where y is uniformly distributed on (0,1) because it is a cdf. Then x is exponentially distributed.

• This method can be used for any distribution in theory. But it is particularly useful for random variates that their inverse function can be easily solved.

• Steps involved are as follows.

  Step 1.

  Compute the cdf of the desired random variable $X$.

  For the exponential distribution, the cdf is $F(x) = 1 - e^{-\lambda x}$.

  Step 2.

  Set R = F(X) on the range of $X$.

  For the exponential distribution, $R = 1 - e^{-\lambda x}$ on the range of $x \ge 0$.

  Step 3.

  Solve the equation F(X) = R for $X$ in terms of $R$.

  For the exponential distribution, the solution proceeds as follows.
  

  $$\begin{array}{ll} 1 - e^{-\lambda X} & = R \\ e^{-\lambda ... ...- R) \\ X & = \frac{-1}{\lambda} ln (1 - R) \\ \end{array}$$

  
  

  Step 4.

  Generate (as needed) uniform random numbers $R_1, R_2, ...$ and compute the desired random variates by
  

  $$X_i = F^{-1} (R_i)$$

  
  

  In the case of exponential distribution
  

  $$X_i = \frac{-1}{\lambda} ln (1 - R_i)$$

  
  

  for i = 1, 2, 3, ... where $R_i$ is a uniformly distributed random number on (0,1).

  In practice, since both $R_i$ AND $1-R_i$ are uniformly distributed random number, so the calculation can be simplified as
  

  $$X_i = \frac{-1}{\lambda} ln~ R_i$$

  
  

• To see why this is correct, recall
  
  $$X_i = F^{-1}(R_i) (*)$$
  
  
  and
  
  $$R_i = F(X_i) (**)$$
  
  

  Because $X_i \le x_0$ is equivelant to $F^{-1}(R_i) \le x_0$, and $F$ is a non-decreasing function (so that if $x \le y$ then $F(x) \le F(y)$) we get $X_i \le x_0$ is equivelant to $F^{-1}(R_i) \le x_0$, which implies that $F (F^{-1}(R_i)) \le F(x_0)$ which is equivelant to $R_i \le F(x_0)$.

  This means
  

  $$P(X_i \le x_0) = P (R_i \le F(x_0))$$
  
  
  Since $R_i$ is uniformly distributed on (0,1) and $F(x_0)$ is the cdf of exponential function which is between 0 and 1, so
  
  $$P(R_i \le F(x_0)) = F(x_0)$$
  
  
  which means
  
  $$P(X_i \le x_0) = F(x_0)$$
  
  
  This says the $F(x_0)$ is the cdf for $X$ and $X$ has the desired distribution.

Once we have this procedure established, we can proceed to solve other similar distribution for which a inverse function is relatively easy to obtain and has a closed formula.