# The Question

Let $$p_X(x)$$ be the pmf of a random variable $$X$$. Find the cdf $$F(x)$$ of $$X$$ and sketch its graph along with that of $$p_X(x)$$ if

a) $$p_X(x) = 1, x = 0$$, zero elsewhere

b) $$p_X(x) = 1/3, x = -1, 0, 1$$, zero elsewhere

c) $$p_X(x) = x/15, x = 1, 2, 3, 4, 5$$, zero elsewhere

# My Attempt

a) $$F_X(x) = 1$$

b) $$F_X(x) = 1/3$$

c) $$F_X(x) = 1/15 + 2/15 + 3/15 + 4/15 + 5/15 = 1$$

I'm unsure which, if any of these are correct.

• These are only valid for x greater than or equal to your rightmost (greatest) values of x. These functions should be piecewise. – David Reed Sep 22 '18 at 23:16
• Could you give me a bit more of a hint? If I at least knew the final answer I could figure out how to get there, but the book doesn't provide one. Am I just making this way more complicated than it needs to be? – John Sep 22 '18 at 23:32
• Sure. See below. – David Reed Sep 22 '18 at 23:42
• Nvm about the delete. Was thinking pdf instead of pmf. My initial answer was correct. – David Reed Sep 22 '18 at 23:58

You need to integrate over the entire region. The CDF of $$x$$ is the sum of all probabilities given by the PMF that are less than $$x$$. Note how in problem B the PMF has a value of $$\frac{1}{3}$$ for each of the 3 corresponding values of $$x$$. Also as a sanity check the CMF should always equal 1 once it is higher than all possible values of x.

A) $$F_X(x)= \begin{cases} 0&\text{if}\, x< 0\\ 1&\text{otherwise} \end{cases}$$

B) $$F_X(x)= \begin{cases} 0&\text{if}\, x< -1\\ \frac{1}{3}&\text{if}\, -1 \leq x < 0\\ \frac{2}{3}&\text{if}\, 0 \leq x < 1\\ 1&\text{otherwise} \end{cases}$$

C) $$F_X(x)= \begin{cases} 0&\text{if}\, x< 1\\ \frac{1}{15}&\text{if}\, 1 \leq x < 2\\ \frac{3}{15}&\text{if}\, 2 \leq x < 3\\ \frac{6}{15}&\text{if}\, 3 \leq x < 4\\ \frac{10}{15}&\text{if}\, 4 \leq x < 5\\ 1&\text{otherwise} \end{cases}$$

I will use $$f(x)$$ for pmf and $$F(x)$$ for cdf. I will show you how to do part b. See if you can figure out the rest.

For discrete distributions the general formula is:

$$F(x) = \sum_{x_i \leq x}f(x_i)$$

As such, each interval $$[x_i,x_{ i+1})$$ should (in general) carry a different value:

For b., we have $$x_1=-1,x_2=0,x_3=1$$

For $$x \in (-\infty,-1)$$ we have not yet hit any $$x_i$$ so $$F(x) = 0$$

For $$x \in [-1,0)$$, $$F(x) = \sum_{x_i \leq x}f(x_i) = f(x_1) = 1/3$$

For $$x \in [0,1)$$, $$F(x) = \sum_{x_i \leq x}f(x_i) = f(x_1)+f(x_2) = 1/3+1/3 = 2/3$$

For $$x \in [1,\infty)$$, $$F(x) = \sum_{x_i \leq x}f(x_i) = f(x_1)+f(x_2)+f(x_3) = 2/3 + 1/3 = 1$$

So altogether,

$$F(x) = \begin{cases} 0 & x < -1 \\ 1/3 & -1 \leq x < 0 \\ 2/3 & 0 \leq x < 1 \\ 1 & x \geq 1 \end{cases}$$