S1 - Discrete random variables again

Hi,

I'm stuck on this last part of this question as I don't really understand it and therefore don't know what to do. I hope someone can explain it to me.

The discrete random variable R has probability function p(r)=P(R=r) defined by:

p(0)=p(6)=1/16, p(1)=p(4)=1/4, p(3)=3/8, p(r)=0 elsewhere

I drew out this frequency distribution to find the mean and variance of R

r P(R=r)
---------
0 1/16

1 1/4

2 0

3 3/8

4 1/4

5 0

6 1/16

I found E(R)=11/4 and Var(R)=37/16, both of which are correct.

Find the mean and variance of:

R1 - R2, where R1 and R2 are independent observations of R. The answers are E(R1 - R2)=0, Var(R1 - R2)=37/8, although I don't understand how you do this question.

Thank you.

Reply 1

silent ninja

well for the last bit

E(R1-R2)= E(R1) - E(R2) using E(aX+b)= aE(x) +b, but since E(R1)=E(R2)= E(R) then
= E(R) - E(R)
= 0

The second part,

Var (R1-R2)= Var(1xR1 + (-1)xR2) but since Var(aX)= a²Var(X) then
= (1)²Var(R1) + (-1)²Var(R2)
= Var(R1) + Var(R2)
= 2 x Var(R)
= 2 x 37/16
= 37/8

basically when you take stuff out the brackets in Var(R1-R2) or E(R1-R2) you have to apply the rules.

Reply 2

sweet_gurl

Thanks for your reply, I understand most of it now. I'm just unsure why R1=R2=R? How do we know this from the question?

Thank you.

Reply 3

silent ninja

Well R1 and R2 have exactly the same probablity distribution as R, as they are just two separate observations of R. For example, if the above distribution was for an event and you carried it out twice and the first outcome is called R1 and the second one R2. The probablity distribution of R1 and R2 separately, are identicle.

The expectation of your first result (R1) minus the second result (R2) is what they require, and then the variance.
You can see why the expectation will be zero. If you think of a die, the mean outcome is 3.5. This will be the same for the first throw, as well as the second throw. So expectation of (first thrown minus second throw) should be zero-- which is exactly what it is. The variance is similar as your results can vary twice as much.

To check the result for the above question, you can do another probablitiy distribution for R1-R2 listing all the possible outcomes (lowest outcome -6, highest 6) and their corresponding probablities, then find the expectation and variance. But if you apply the rules you will get the same answer.
(BTW the explanations in S1 are close to useless).