Question

2. Let X be a continuous random variable. Let R be the set of all real numbers, let Z be the set of all integers, and let Q be the set of all rational numbers. Please calculate (1) P(X ? R), (2) P(X ? Z), and (3) P(X EQ)

Answer 1

2

Since X is a continuous random variable, probability of a point is zero and only intervals have non zero probabilities. Therefore

(1) $P(X\in R)=1$ since the support of X will be a subset of R.

(2)

. $P(X\in Z)=\sum_{i\in Z}P(X=i)=0$ since probability of a point is zero for continuors random variable.

(3) It may be noted that set of all rational numbers is a countable set.

Therefore

$P(X\in Q)=\sum_{i\in Q}P(X=i)=0$

(6)

Given $f(x)=\frac{\lambda}{2}e^{-\lambda |x|}$

By definition the distribution function is given by

$F(x)=\int_{-\infty}^xf(y)dy.$

Since the deinsity function has |x| term we will calculate the distribution function for two ranges separately, that is wehn x<=0 and when x>0.

When x<=0, |x|= - x and

$F(x)=\int_{-\infty}^xf(y)dy=\int_{-\infty}^x\frac{\lambda}{2}e^{-\lambda \times -y}dy\\ =\frac{\lambda}{2}\int_{-\infty}^xe^{\lambda y}dy\\ =\frac{\lambda}{2}\frac{e^{\lambda y}}{\lambda}|_{-\infty}^x\\ =\frac{e^{\lambda x}}{2}$

When x>0.

$F(x)=\int_{-\infty}^xf(y)dy=\int_{-\infty}^0f(y)dy+\int_{0}^xf(y)dy\\ =\frac{1}{2}+\int_{0}^x\frac{\lambda}{2}e^{-\lambda y}dy\\ =\frac{1}{2}+\frac{\lambda}{2}\frac{e^{-\lambda y}}{-\lambda}|_{0}^x\\ =\frac{1}{2}+\left(-\frac{e^{-\lambda x}}{2}+\frac{1}{2} \right )\\ =1-\frac{e^{-\lambda x}}{2}$

Therefore the distribution function is

$F(x)= \left\{\begin{matrix} \frac{e^{\lambda x}}{2} \quad if \quad x \le0 \\ 1-\frac{e^{-\lambda x}}{2} \quad if \quad x>0 \end{matrix}\right.$