Q. 10.8 - A First Course in Probability

Question

Suppose it is relatively easy to simulate from Fi for each i = 1, ... , n. How can we simulate from 

(a) $F\left(x\right)=\prod _{i=1}^n{F}_i\left(x\right) ?$

(b)$F\left(x\right)=1-\prod _{i=1}^n\left[1-F_i\left(x\right)\right] ?$

Step-by-Step Solution

Answer

(a) The CDF is the CDF of maximum.

(b) The CDF is the CDF of minimum.

Part (a) Step 1: Given Information

We have to prove the statement

$F\left(x\right)=\prod _{i=1}^n{F}_i\left(x\right)$

Part (a) Step 2: Simplify

Suppose that  follows the distribution with CDF $F_i,i=1,\dots ,n$.
 (a)
Observe that $F$ is the CDF of the $\text{max}\left(X_1,\dots ,X_n\right)$. Indeed

$\begin{array}{l}F\left(x\right)=P\left(m \left(X_1,\dots ,X_n\right)\le x\right)=P\left(X_1\le x,\dots ,X_n\le x\right)\\ =\prod _{i=1}^{n}P\left(X_i\le x\right)=\prod _{i=1}^n{F}_i\left(x\right)\end{array}$

So, generating$X_1,...,X_n$ considering its maximum, call it X. From the fact shown above, we have $X$ that has required CDF.

Part (b) Step 1: Given Information

We have to prove the statement

$F\left(x\right)=1-\prod _{i=1}^n\left[1-F_i\left(x\right)\right]$

Part (b) Step 2: Simplify

(b)

Observe that $F$ is the CDF of the $\text{min}\left(X_1,\dots ,X_n\right)$. Indeed

$$\begin{gathered} 1-F\left(x\right) =P\left(m \left(X_1,\dots ,X_n\right)>x\right)=P\left(X_1>x,\dots ,X_n>x\right) \\ =\prod _{i=1}^{n}P(X_i>x)=\prod _{i=1}^n(1-F_i(x\left)\right) \end{gathered}$$

which implies 

$F\left(x\right)=1-\prod _{i=1}^n\left(1-F_i\left(x\right)\right)$

So, generating$X_1,.....,X_n$and consider its minimum, call it $X$. From the fact shown above, we have$X$ has required CDF.