Question

2.34. Probability integral transformation. Consider a random variable X with cumulative function Fx(x), 0-x-00, Now define a new random variable U to be a particular function of X, namely, U = Fx(X) For example, if FX(x)-1-e-Ax, then U = 1-e-Ax = g(X). Show [at least for reasonably smooth Fx(x)] that the random variable U has a constant density function on the interval O to 1 and is zero elsewhere. Hint: Con vince yourself graphically thatgg (u)- u and assume that Fx(x) satis- fies the conditions needed to apply Eq. (2.3.3) To illustrate this notion, sketch the Fx(x) and F(u)for several CDF's, including the case where Xis a mixed random variable and the case where Fx(x) equals a constant, say 0.5, over an interval. [These cases sug- gest why the definition of the inverse function Fx (u)is better stated as "smallest value of x for which Fx) is greater than or equal to u. With this definition of the inverse function, Eg: (2 : 3 : 3 ) will hold for any practical CDF.] The practical importance of this, the "probability integral transfor- mation," lies in the fact that in simulation studies (Sec 2.3.3) one can sample the variable X by the easier process of first sampling the variable U, finding the value u, and then calculating the corresponding value of X by If Fx (u) cannot be evaluated explicitly, the finding of x given u can always be done graphically on a plot of Fx(x) (or in a computer by "table lookup") For the example distribution above (withA - 1), find the values of the random variable X corresponding to the random numbers 0.51, 0.20 and 0.98 drawn from a table. These are observations of U.

Answer 1

For a reasonably smooth F_X(x) the random variable U = F_X(X) has a constant density function on [0,1] and zero elsewhere

Proof:

Let U = F_X(X) for a reasonable smooth F_X(x).

Then the cdf of U is given by:
$\begin{align*} F_U(u) &= P(U \le u) \ \ \ \ \ \ \ \ \ \ \ \ ; 0 \le u \le 1 \\ &= P(F_X(X) \le u) \ \ \ \ \ ; 0 \le u \le 1 \\ &= P(X \le F^{-1}_X(u)) \ \ \ \ \ ; 0 \le u \le 1 \\ &= F_X(F^{-1}_X(u)) \ \ \ \ \ \ \ \ \ \ ; 0 \le u \le 1 \\ &= u \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ; 0 \le u \le 1 \end{align*}$

Thus, the pdf of X is given by:

$\begin{align*} f_U(u) &= \frac{\mathrm{d} }{\mathrm{d} u} F_U(u) \ \ \ \ \ ; 0 \le u \le 1 \\ &= \frac{\mathrm{d} }{\mathrm{d} u} (u) \ \ \ \ \ ; 0 \le u \le 1 \\ &= 1 \ \ \ \ \ ; 0 \le u \le 1 \\ &= \begin{cases} 1 & \text{ if } 0 \le u \le 1 \\ 0 & \text{ otherwise } \end{cases} \end{align*}$

For the distribution above (with $\lambda$ =1), find the values of random variable X...

We are given that X ~ Exponential($\lambda$ = 1)

Thus,

$\begin{align*} U &= F_X(X) \\ \Rightarrow U &= 1 - e^{-X} \\ \Rightarrow e^{-X} &= 1 -U \\ \Rightarrow X &= -ln(1-U) \end{align*}$

Putting the values of U given in question in the equation above, we get:

$\begin{align*} \text{For U = 0.51} \ \ \ \ \Rightarrow X &= -ln(1-0.51) = 0.713350 \\ \text{For U = 0.20} \ \ \ \ \Rightarrow X &= -ln(1-0.20) = 0.223144 \\ \text{For U = 0.98} \ \ \ \ \Rightarrow X &= -ln(1-0.98) = 3.912023 \end{align*}$

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