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This question for my homework I am truly lost on, I have some idea for how to solve part of it but I am ultimately unsure where to start.

Consider pmf px for a random variable X: p(1)=1/2; P(3)=1/4; P(6)=1/4

a. Draw the CDF Fx
b. What is E|X|?
c. If g(x) = e^[(x-2)^2], what is E[g(X)]?

I have been tearing through my book to try and gain some understanding but i am not having much luck. Any help on how to start this problem would be greatly appreciated.

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    Admitting right away that it is a homework problem. I like you. This sounds sarcastic, but I assure you it is not. – 2017-02-27

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Some hints that hopefully can help you get started:

(a) $$F_X(1) = Pr(X \leq 1)=Pr(X=1)$$ $$F_X(2.5) = Pr(X \leq 2.5)=Pr(X=1)$$ $$F_X(3) = Pr(X \leq 3)=Pr(X=1)+Pr(X=3)$$

Your answer should look like a step function.

(b) $$\mathbb{E}[X] = \sum_x xp(x)$$ (c) $$\mathbb{E}[g(X)] = \sum_x g(x)p(x)$$

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    I think my main understanding is that I am imagining values for the pmf that are not stated - is the general idea in this sort of question that the pmf is given is all there is? Specifically referring to the given p(1)=1/2; P(3)=1/4; P(6)=1/4. If this were the case, then I know I can calculate b) as E|X| = (1)(1/2)+(3)(1/4)+(6)(1/4) = 11/4 – 2017-02-27
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    For this question, $p(1)+p(3)+p(6)=1$. Since probabilities are nonnegative and sum to 1, $p(x)=0$ for $x \notin \{ 1,3,6\}$. – 2017-02-27
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    If the sum of the probability masses you have equals one, then you have all the probability masses there are. @Winguh – 2017-02-27
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    That is something very simple I overlooked that explains a lot more, thank you for this clarification – 2017-02-27
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    Alright given what I have learned so far I have found b) 11/4 and c) (3e+e^16)/4. As I understand a, p(x) will be 1/2 when x is from 0 to 1, 3/4 when x is from 1 to 3, and 1 from 3 to 6? – 2017-02-27
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    b and c seems fine. part a is a bit strange. In particular $F_X(0.5)=0$, $F_X(4)<1$. – 2017-02-27
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    [This](http://imgur.com/a/uJw7F) is a very poorly drawn example of what I think is the correct approach – 2017-02-27
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    seems fine. Remember to define your function for $x \in [2,3)$ as well. – 2017-02-27
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    Ah yes I see I left out a part - so does the top "step" continue to infinity, or is it just a single point? – 2017-02-27
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    $F_X(x) = 1$ for $x \geq 6$. Proof: If $x \geq 6$, then we have $y \leq 6 \implies y \leq x$. $1=Pr(X \leq 6) \leq Pr(X \leq x)=F_X(x)$. – 2017-02-27