# Math Help - Simple statistics

1. ## Simple statistics

if E(X) = 1.6 and E(Y) = 2

does E (Y/X) = E(Y)/E(X)

2/1.6 ??

2. Originally Posted by djmccabie
if E(X) = 1.6 and E(Y) = 2

does E (Y/X) = E(Y)/E(X)

2/1.6 ??
No. It might be best if you post the entire question.

3. The random variables X and Y are independent and have the following probability distributions.

x 1 2
P(X=x) 0.4 0.6

Y 1 2 3
P(Y=y) 0.3 0.4 0.3

i) Evaluate E(Y/X)

4. Originally Posted by djmccabie
The random variables X and Y are independent and have the following probability distributions.

x 1 2
P(X=x) 0.4 0.6

Y 1 2 3
P(Y=y) 0.3 0.4 0.3

i) Evaluate E(Y/X)
$
E(Y/X)= 1 \times 0.4 \times 0.3 + 2 \times 0.4 \times 0.4 + 3\times 0.4 \times 0.3 +$

............... $\
(1/2) \times 0.6 \times 0.3+(2/2) \times 0.6 \times 0.4+(3/2) \times 0.6 \times 0.3
$

CB

5. i dont understand any of that

6. Originally Posted by djmccabie
The random variables X and Y are independent and have the following probability distributions.

x 1 2
P(X=x) 0.4 0.6

Y 1 2 3
P(Y=y) 0.3 0.4 0.3

i) Evaluate E(Y/X)
Originally Posted by CaptainBlack
$
E(Y/X)= 1 \times 0.4 \times 0.3 + 2 \times 0.4 \times 0.4 + 3\times 0.4 \times 0.3 +$

............... $\
(1/2) \times 0.6 \times 0.3+(2/2) \times 0.6 \times 0.4+(3/2) \times 0.6 \times 0.3
$

CB
Originally Posted by djmccabie
i dont understand any of that
That doesn't bode well.

What are the possible values of Y/X ......? 1/1 = 1, 1/2, 2/1 = 2, 2/2 = 1 (already considered), 3/1 = 3, 3/2.

Pr(Y/X = 1) = Pr(Y = 1 and X = 1) + Pr(Y = 2 and X = 2) = (0.3)(0.4) + (0.4)(0.6) = 0.12 + 0.24 = 0.36.

Pr(Y/X = 1/2) = Pr(Y = 1 and X = 2) = (0.3)(0.6) = 0.18.

etc.

Then E(Y/X) = (1)(0.36) + (1/2)(0.18) + ......

This is the same calculation as CaptainB did - all I've done is simplified the arithmetic.

7. Do you uderstand what P(Y/X) means? Remember that Y and X are events not numbers. What is meant by Y/X?

8. Originally Posted by HallsofIvy
Do you uderstand what P(Y/X) means? Remember that Y and X are events not numbers. What is meant by Y/X?
I'm not sure who your post is addressed is to. Nevertheless, X and Y are not events, they are random variables and have a probability of assuming particular values.

9. Originally Posted by djmccabie
i dont understand any of that
It uses the definition of expectation for a discrete random variable:

$E(U)=\sum_i u_i p(u_i)$

where the $u_i$'s are the possible values of $U$, and $p(u_i)$ are the probabilities of those values.

Also it uses the definition of independent RV that $p(a,b)=p(a)p(b).$

So in this case we have:

$E(Y/X)=\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i,y=j) =\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i)p(y=j)$

CB

10. Originally Posted by HallsofIvy
Do you uderstand what P(Y/X) means? Remember that Y and X are events not numbers. What is meant by Y/X?
They are random variables, not events. We are finding expectations, which are properties of RV.

Also no one has used P(Y/X) as a notation (which if they had might be mistaken for the conditional probability P(Y|X) but neither I nor MrF would do that -would we?)

Also $^2$, please quote what you are replying to so we know what you are talking about (and if you don't know what you are talking about don't talk about it).

CB

11. Originally Posted by CaptainBlack
It uses the definition of expectation for a discrete random variable:

$E(U)=\sum_i u_i p(u_i)$

where the $u_i$'s are the possible values of $U$, and $p(u_i)$ are the probabilities of those values.

Also it uses the definition of independent RV that $p(a,b)=p(a)p(b).$

So in this case we have:

$E(Y/X)=\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i,y=j) =\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i)p(y=j)$

CB

Not 1 for simplicity are you? lol

Its ok tho, i know how to work it out now.

Ill ask my tutor why when college starts again