Functions of a random variable  (Page 2/3)

A discrete example

Suppose X has values -2, 0, 1, 3, 6, with respective probabilities 0.2, 0.1, 0.2, 0.3 0.2.

Consider $Z=g\left(X\right)=(X+1)(X-4)$ . Determine $P(Z>0)$ .

SOLUTION

First solution . The mapping approach

$g\left(t\right)=(t+1)(t-4)$ . $N=\{t:g(t)>0\}$ is the set of points to the left of $-1$ or to the right of 4. The X -values $-2$ and 6 lie in this set. Hence

Second solution . The discrete alternative

Picking out and adding the indicated probabilities, we have

In this case (and often for “hand calculations”) the mapping approach requires less calculation. However, for MATLAB calculations (as we show below), the discrete alternative is more readily implemented.

An absolutely continuous example

Suppose $X\sim$ uniform $[-3,7]$ . Then $f_X\left(t\right)=0.1,-3\le t\le 7$ (and zero elsewhere). Let

Determine $P(Z>0)$ .

SOLUTION

First we determine $N=\{t:g(t)>0\}$ . As in [link] , $g\left(t\right)=(t+1)(t-4)>0$ for $t<-1$ or $t>4$ . Because of the uniform distribution, the integral of the density over any subinterval of $[-3,7]$ is 0.1 times the length of that subinterval. Thus, the desired probability is

The normal distribution and standardized normal distribution

To show that if $X\sim N(\mu ,{\sigma }^2)$ then

VERIFICATION

We wish to show the denity function for Z is

Now

Hence, for given $M=(-\infty ,v]$ the inverse image is $N=(-\infty ,\sigma v+\mu ]$ , so that

Since the density is the derivative of the distribution function,

Thus

We conclude that $Z\sim N(0,1)$

Affine functions

Suppose X has distribution function F _X . If it is absolutely continuous, the corresponding density is f _X . Consider $Z=aX+b(a\ne 0)$ . Here $g\left(t\right)=at+b$ , an affine function (linear plus a constant). Determine the distribution function for Z (and the density in the absolutely continuous case).

SOLUTION

There are two cases

• $a>0$ :
  $$F_Z\left(v\right)=P\left(X,\le ,\frac{v-b}{a}\right)=F_X\left(\frac{v-b}{a}\right)$$
• $a<0$
  $$F_Z\left(v\right)=P\left(X,\ge ,\frac{v-b}{a}\right)=P\left(X,>,\frac{v-b}{a}\right)+P\left(X,=,\frac{v-b}{a}\right)$$
  So that
  $$F_Z\left(v\right)=1-F_X\left(\frac{v-b}{a}\right)+P\left(X,=,\frac{v-b}{a}\right)$$

For the absolutely continuous case, ${\displaystyle P\left(X,=,\frac{v-b}{a}\right)=0}$ , and by differentiation

• for ${\displaystyle a>0f_Z\left(v\right)=\frac{1}{a}{f}_X\left(\frac{v-b}{a}\right)}$
• for ${\displaystyle a<0f_Z\left(v\right)=-\frac{1}{a}{f}_X\left(\frac{v-b}{a}\right)}$

Since for $a<0$ , $-a=\left|a\right|$ , the two cases may be combined into one formula.

Completion of normal and standardized normal relationship

Suppose $Z\sim N(0,1)$ . Show that $X=\sigma Z+\mu (\sigma >0)$ is $N(\mu ,{\sigma }^2)$ .

VERIFICATION

Use of the result of [link] on affine functions shows that

Fractional power of a nonnegative random variable

Suppose $X\ge 0$ and $Z=g\left(X\right)=X^{1/a}$ for $a>1$ . Since for $t\ge 0$ , $t^{1/a}$ is increasing, we have $0\le t^{1/a}\le v$ iff $0\le t\le v^a$ . Thus

In the absolutely continuous case

Fractional power of an exponentially distributed random variable

Suppose $X\sim$ exponential $\left(\lambda \right)$ . Then $Z=X^{1/a}\sim$ Weibull $(a,\lambda ,0)$ .

According to the result of [link] ,

which is the distribution function for $Z\sim$ Weibull $(a,\lambda ,0)$ .

A simple approximation as a function of X

If X is a random variable, a simple function approximation may be constructed (see Distribution Approximations). We limit our discussion to the bounded case, in which the rangeof X is limited to a bounded interval $I=[a,b]$ . Suppose I is partitioned into n subintervals by points t _i , $1\le i\le n-1$ , with $a=t_0$ and $b=t_n$ . Let $M_i=[t_{i-1},t_i)$ be the i th subinterval, $1\le i\le n-1$ and $M_n=[t_{n-1},t_n]$ . Let $E_i=X^{-1}\left(M_i\right)$ be the set of points mapped into M _i by X . Then the E _i form a partition of the basic space Ω . For the given subdivision, we form a simple random variable X _s as follows. In each subinterval, pick a point $s_i,t_{i-1}\le s_i<t_i$ . The simple random variable

approximates X to within the length of the largest subinterval M _i . Now $I_{E_i}=I_{M_i}\left(X\right)$ , since $I_{E_i}\left(\omega \right)=1$ iff $X\left(\omega \right)\in M_i$ iff $I_{M_i}\left(X\left(\omega \right)\right)=1$ . We may thus write