Question

6. (20 points) Let X be a continuous random variable such that, P(X <0) equals zero and for each 0, the cumulative density function Fx) is differentiable and its derivative equals the probability density function fx (a) (6 points) Given 0 < b< a, find an expression of the probability P(b X a) in terms of the function FX (-), what is the value of FX(0)? b) (7 points) Let a > 0. Use integration by parts to show that C1 0 c) (7 points) Using the above, show that Oo E(X) =「P(X > z)dz. J0

I am not familiar with this kind of questions. Please help me solve this. Your work will be greatly appreciated. THX !!

Answer 1

Since X is a continuous random variable and P(X<0) = 0. We have

$f_X(x) = \frac{\mathrm{d}(F_X(x)) }{\mathrm{d} x}\Rightarrow P(X\leq x) = F(x) = \int_{0}^{x}f(x)dx$

a) Since 0 < b < a. Now

$P(a \leq X\leq b) = \int_{a}^{b}f(x)dx = \int_{a}^{b}\frac{\mathrm{d} F(x)}{\mathrm{d} x}dx=\\ \int_{a}^{b}\mathrm{d} F(x) = F(b)-F(a)$

$F_X(0) = P(X\leq 0) = 0\text{ (given in question)}$

b) a > 0.

$\int_{0}^{a}xf_X(x)dx = x\int_{0}^{a}f_X(x)dx-\int_{0}^{a}[\frac{\mathrm{d} (x)}{\mathrm{d} x}\int f_X(x)dx]dx=\\ xF_X(x)|_{0}^{a} - \int_{0}^{a} F_X(x)dx = (aF_X(a)-\lim_{x\rightarrow 0}xF_X(x)) - \int_{0}^{a} F_X(x)dx.\\ Now,\: \: \lim_{x\rightarrow 0}xF_X(x) = \text{(By L'Hospital Rule)} \lim_{x\rightarrow 0}f_X(x) = 0.\\ Therefore,\: \: \int_{0}^{a}xf_X(x)dx =aF_X(a) - \int_{0}^{a} F_X(x)dx= \int_{0}^{a} F_X(a)dx-\int_{0}^{a} F_X(x)dx =\\ \int_{0}^{a} (F_X(a) - F_X(x))dx$

c)

$E[X] = \int_{0}^{\infty}xf_X(x)dx= \int_{0}^{\infty} (F_X(\infty) - F_X(x))dx.\\ Now, \: \: \lim_{x \rightarrow \infty}F_X(x) = F_X(\infty) = P(X\leq \infty) = 1(\text{ The number is always less than infinity} ).\\ Therefore, \: \: E[X] = \int_{0}^{\infty} (1 - F_X(x))dx.\\ Also, \: \: P(X > x) = 1-P(X\leq x) = 1 - F_X(x). \\ Hence, \: \: E[X] = \int_{0}^{\infty} P(X>x)dx$