Q:

Expectation Help

I am having trouble a question in my probability course and was looking for some insight into it.

A digital clock sits next to your bed and in the mornings when you don't use an alarm you notice Y = the last digit of the time you wake up (e.g. if you wake up at 7:18 am then Y = 8).
a) find E(Y+1) and E[(Y+1)^2]?
b) find E[1/(Y+1)]

Would I just add 1 to each of the possible Y variables and multiply them like normal to the probability of it being a certain minute? For example if the minute was 1, I would + 1 so it would be 2(1/10)?
Any Help please?

A:

Some hints, happy to expand further or write out full solution if helpful:
Re the comments, your E[Y] is incorrect. Recall that in the discrete case, expectation is defined as 
"Let $W$ be a random variable with a finite number of finite outcomes $\{w_1,w_2,...,w_k\}$ with probabilities $\{p_1,...,p_k\}$ such that $\sum_{i=1}^k p_i = 1$. Then the expectation of $W$ is $$E[W] = w_1p_1+...+w_kp_k$$"
So let's use this to calculate E[Y]. What are the values of Y? 0,1,2,3,4,5,6,7,8,9. What is the probability of each one occurring? Well, they all have an equal likelihood of happening (make sure this makes sense to you), so the probability of any of them happening is 1/(number of possibile values) = 1/10 =.1. So going back to formula for expected value, $$E[Y] = (0)(.1)+(1)(.1)+...+(9)(.1) = 4.5$$ Another simpler way is to realize that $p_k = p =.1$ for all values, so $$E[Y] = \sum_{k=0}^9 w_kp_k = \sum_{k=0}^9 w_k(.1) = (.1)\sum_{k=0}^9 k = 4.5$$ because w_0=0, w_1=1, etc. Does this make sense?
Overall, think about what information is important to the question and what isn't. For example, do you care at all what the values that are not the last value? Is 8:03 any different than 7:13 in this case? So what are the possible values of Y, and what is the average value? 
for problem a), what you wrote is basically correct. Recall linearity of expectation operator:
$$E[aX+bY] = aE[X]+b[Y]$$ where a,b are constants and X and Y are random variables. Using this, what would $E[aY+b]$ be? This should help with the first question. Also remember that you can always expand the inside of an expectation and see if that leads anywhere bearing in mind this linearity :). 
for problem b), recall the definition of expected value when you have finite outcomes:
Let $W$ be a random variable with a finite number of finite outcomes $\{w_1,w_2,...,w_k\}$ with probabilities $\{p_1,...,p_k\}$ such that $\sum_{i=1}^k p_i = 1$. Then the expectation of $X$ is $$E[Y] = w_1p_1+...+w_kp_k$$ Maybe use this for problem 2 and see where it goes? Maybe try defining a variable W where W = 1/(Y+1), and for each Y=y, what is W? and then maybe use this definition, given that you know the probability of each value? Here's a simple example to make what is going on clear: let $T=4$ with probability $1/2$, and let $T=6$ with probability $1/2$. Then $$E(T)= (4)(1/2) + (6)(1/2) = 5$$ but $$E(1/T)=(1/4)(1/2)+(1/6)(1/2) \approx .208$$
Hope this helps!

