Now we will give some problems related to Cumulative distribution function so you can have a deeper understanding of this topic. These problems are adapted from the following textbook:

“Probability and Random Processes by Scott Miller 2nd Edition.”

Problem-1: Which of the following mathematical functions could be the CDF of some random variable?

Remember: To be a CDF of some random variable, the function $$F_{X}(x)$$ must start at zero when $$x = -\infty$$, end at one when $$x = \infty$$, and be monotonic increasing in between

$$1- \begin{eqnarray*} F_X(x) &=& 1-|x|, & |x| \leq 1,\\ & & 0 , & |x| <1 \end{eqnarray*}$$

Answer: This cannot be the CDF of a random variable since the function is not monotonic increasing

$$2- \begin{eqnarray*} F_X(x) &=& \frac{1}{2} e^x, & x < 0,\\ & & 1 – \frac{1}{2} e^{-x} , & x \geq 0 \end{eqnarray*}$$

Answer: $$F_{X}(x)$$ satisfies the necessary conditions, and thus could be the CDF of some random variable.

$$3- \begin{eqnarray*} F_X(x) &=& 0, & x \leq 0,\\ & & x^2 , & 0 < x \leq 1 \\ & & 1 , & x > 1 \end{eqnarray*}$$

Answer: $$F_{X}(x)$$ satisfies the necessary conditions, and thus could be the CDF of some random variable

$$4- F_X(x) = log(x)u(x)$$

Answer: This function cannot be the CDF of some random function. Taking the limit as $$x \rightarrow \infty$$ we see the function will grow without bound, so it will not end at one as required.

Problem-2: Suppose a random variable has a CDF given by:

$$\begin{equation*} F_X(x) = \frac{x^2}{1+x^2} u(x) \end{equation*}$$

Find the following quantities:

(a) $$Pr(X<2)$$
(b) $$Pr(X>4)$$
(c) $$Pr(1 < X < 3)$$
(d) $$Pr(X>2 | X < 4)$$

Solution: (a) $$Pr(X<2) = F_{x}(2) = \frac{2^{2}}{1+2^{2}} = 4/5$$
(b) $$Pr(X>4) = 1-Pr(X\leq4) = 1 – F_{x}(4) = 1 – \frac{4^{2}}{1+4^{2}} = 1/17$$
(c) $$Pr(1<X<3) = Pr(X<3) – Pr(X<1) = Pr(X<3) – Pr(X<1) = F_{x}(3) – F_{x}(1) \\ = \frac{3^{2}}{1+3^{2}} – \frac{1^{2}}{1+1^{2}} = 2/5$$

(d)  $$Pr(X>2 | X<4) = \frac{\text{Pr}\bigg((X>2)\cap(X<4)\bigg)}{\text{Pr}(X<4)} \\ =\frac{\text{Pr}(2<X<4)}{\text{Pr}(X<4)}=\frac{F_{x}(4)-F_{x}(2)}{F_{x}(4)}} \\ = {\frac{\frac{4^{2}}{1+4^{2}}-\frac{2^{2}}{1+2^{2}}}{\frac{4^{2}}{1+4^{2}}}}} \\ = 3/20 = 0.1$$

Problem-3: Now Suppose a random variable has a CDF given by:

$$\begin{equation*} F_X(x) = \left( \frac{1}{2} + \frac{1}{\pi} tan^{-1}(x) \right) u(x)\end{equation*}$$

Find the following quantities:

(a) $$Pr(X<2)$$
(b) $$Pr(X>4)$$
(c) $$Pr(1 < X < 3)$$
(d) $$Pr(X>2 | X < 4)$$

Solution: (a) $$Pr(X<2) = F_{x}(2) = \big(\frac{1}{2} + \frac{1}{\pi} tan^{-1}(2) \big). \\$$
(b) $$Pr(X>4) = 1-Pr(X \leq 4) = 1 – F_{x}(4) \\ = 1 – \big(\frac{1}{2} + \frac{1}{\pi} tan^{-1}(4) \big) \\ = \big(\frac{1}{2} – \frac{1}{\pi} tan^{-1}(4) \big). \\$$
(c) $$Pr(1<X<3) = Pr(X<3) – Pr(X<1) = F_{x}(3) – F_{x}(1) \\ = \big(\frac{1}{2} + \frac{1}{\pi} tan^{-1}(3) \big) -\big(\frac{1}{2} + \frac{1}{\pi} tan^{-1}(1) \big) \\ = \frac{1}{\pi} tan^{-1}(3) – \frac{1}{\pi} tan^{-1}(1) \\$$
Since $$tan^{-1}(1) = \frac{\pi}{4}$$, we have
$$Pr(1<X<3)=-\frac{1}{4}+\frac{1}{\pi}\tan^{-1}(3). \\$$
(d) $$Pr(X>2 | X<4)$$
\begin{align*} } & =\frac{\text{Pr}\bigg((X>2)\cap(X<4)\bigg)}{\text{Pr}(X<4)}=\frac{\text{Pr}(2<X<4)}{\text{Pr}(X<4)}=\frac{F_{x}(4)-F_{x}(2)}{F_{x}(4)}\\ & =\frac{\big(\frac{1}{2}+\frac{1}{\pi}\tan^{-1}(4)\big)-\big(\frac{1}{2}+\frac{1}{\pi}\tan^{-1}(2)\big)}{\big(\frac{1}{2}+\frac{1}{\pi}\tan^{-1}(4)\big)}=\frac{\tan^{-1}(4)-\tan^{-1}(2)}{\frac{\pi}{2}+\tan^{-1}(4)}. \end{align*

Problem-3: Suppose we flip a balanced coin five times and let the random variable $$X$$ represent the

(a) Sketch the CDF of $$X, F_X(x)$$
(b) Write $$F_X(x)$$ analytically in terms of unit step functions.

Solution: (a) Let X be the number of heads in five coin tosses, so X can be 0, 1, 2, 3, 4, or 5.
$$Pr(X=0) = (\frac{1}{2})^{5} = \frac{1}{32}. \\ Pr(X=1) = {5 \choose 1}\big(\frac{1}{2}\big)^{4}\big(\frac{1}{2}\big)^{1} = \frac{5}{32}. \\ Pr(X=2) = {5 \choose 2}\big(\frac{1}{2}\big)^{3}\big(\frac{1}{2}\big)^{2} = \frac{10}{32}. \\ Pr(X=3) = {5 \choose 3}\big(\frac{1}{2}\big)^{2}\big(\frac{1}{2}\big)^{3} = \frac{10}{32}. \\ Pr(X=4) = {5 \choose 4}\big(\frac{1}{2}\big)^{4}\big(\frac{1}{2}\big)^{1} = \frac{5}{32}. \\ Pr(X=5) = {5 \choose 5}\big(\frac{1}{2}\big)^{5} = \frac{1}{32}. \\$$

Now $$F_{x}(x) = {\sum_{i=1}^{k}}} P_{x}(x_{i}), for x_{k}\leq x \leq x_{k+1} \$$

$$F_{x}(x)=\left\{ \begin{array}{ll} 0 & \quad\text{for x < 0}\\ 1/32 & \quad\text{for 0 \leq x < 1}\\ 6/32 & \quad\text{for 1 \leq x < 2}\\ 1/2 & \quad\text{for 2 \leq x < 3}\\ 26/32 & \quad\text{for 3 \leq x < 4}\\ 31/32 & \quad\text{for 4 \leq x < 5}\\ 1 & \quad\text{for 5 \leq x} \end{array}\right.$$ (b) $$F_{x}(x) = \frac{1}{32}u(t) + \frac{5}{32}u(t-1) + \frac{10}{32}u(t-2) + \frac{10}{32}u(t-3) + \frac{5}{32}u(t-4) + \frac{1}{32}u(t-5)$$