Stat question

[UncleWai]

link to pdf

question 4a.
I don't even know where to begin.
Is the expected value the same as sample mean?
for the Variance, I don't know which variables I need to use to derive the answer from.

 

[TuxDave]

E(A+B+C) = E(A)+E(B)+E(C)

E = Expected Value
A,B,C = Independant Random Variables

E(A) = Mean of A? I'm guessing?

Var(A+B+C) = Var(A)+Var(B)+Var(C)?

STDEV = sqrt(VAR)

So far so good?

 

[UncleWai]

I think Y is just a random number.

So all it is asking is generating many different random numbers.
Since it already provides the sample mean of the sample, shouldn't the expected value just be the same?
For variance,
sigma( (Yi - y bar) ^2)

but I don't know how I can simplify any more?

 

[TuxDave]

Originally posted by: UncleWai
I think Y is just a random number.

So all it is asking is generating many different random numbers.
Since it already provides the sample mean of the sample, shouldn't the expected value just be the same?
For variance,
sigma( (Yi - y bar) ^2)

but I don't know how I can simplify any more?

Well... if they're asking for the derivation, you gotta show the steps.

So ybar = 1/n*(y1+y2+y3.....yn)

E(ybar) =
E[1/n*(y1+y2+y3.....yn)] =
1/n*E(y1+y2+y3.....yn) =
1/n*[E(y1)+E(y2)....E(yn)] =
1/n*(mu+mu+mu+mu.....+mu) =
1/n*(n*mu) = mu

I'm guessing that's what they want. You have to do the same for variance.

Var(ybar) =
Var[1/n*(y1+y2+y3.....yn)] =
....

There should be a identity for variance that you can use... something like.

Var(k*(A+B)) = k^2*[Var(A)+Var(B)]

If K is a constant and A and B are independant random variables. Let me know if I'm going too fast.

 

[chuckywang]

Mean of sums of random variables is just the sum of the means of the random variables. However, variance does not have that linearity property.

Var[aX + bY] = a^2*Var[X] + b^2 *Var[Y] if X and Y are independent random variables and a,b are constants. You can easily extend this for more than two random variables.

 

[UncleWai]

i don't get it.
Var(A) Var(B) = (yi - u)^2 right?

so I take out the 1/n which is a constant.

(1/n)^2 * sigma (yi-U)^2 is the answer?


I also got stuck on Q5.

I don't see how I can use the var x, varu, cov xu, E(U) = 0, E(X) = 0, y = x + u to form cov(xy)


Just using symbols to find answer these questions puts statistic into a whole new dimension, I really can't grasp it.

 

[TuxDave]

Originally posted by: UncleWai
i don't get it.
Var(A) Var(B) = (yi - u)^2 right?

so I take out the 1/n which is a constant.

(1/n)^2 * sigma (yi-U)^2 is the answer?


I also got stuck on Q5.

I don't see how I can use the var x, varu, cov xu, E(U) = 0, E(X) = 0, y = x + u to form cov(xy)


Just using symbols to find answer these questions puts statistic into a whole new dimension, I really can't grasp it.

Its given in the problem that Var(yi) = sigma^2

It also gave you ybar = 1/n*(y1+y2+y3...+yn)

It's asking you to find var(ybar) which equals var[1/n*(y1+y2+y3...+yn)]

Use the identities we posted above to evaluate it in terms of sigma. Sigma is just sigma just like how mu is just mu. Sigma is not a function so sigma(x) doesn't make sense.

 

[TuxDave]

As for #5... I think you need to brush up on some statistics equation.

y = x+u

You're given var(x) and var(u)
From problem 4b, you should be able to easily find var(y) in terms of var(X) and var(u)

Cor = Correlation
Cov = Covariance

Here's an equation for correlation.

Cor(x,y)= Cov(x,y)/sqrt(var(x)*var(y))

Here's an equation for covariance.

var(x+y) = var(x)+var(y)+2*cov(x,y)

It's kinda like algebra. You're given all these relationships, and you gotta move from one end to the other. You just got to remember the basic relationships between variance, standard deviation, covariance, mean, etc.....

 

[UncleWai]

I have this portion sqrt(var(x)*var(y)) down.

But I don't know what I can do with Cov(x,y)

 

[TuxDave]

Originally posted by: UncleWai
I have this portion sqrt(var(x)*var(y)) down.

But I don't know what I can do with Cov(x,y)

I'm looking up some identities... umm... I dunno if you can use this, but it says it's an identity... so whatever

Here's your hints....

y=x+u

Cov(a,b+c) = Cov(a,b)+Cov(a,c)

Oh... and remember

Cov(x,x) = Var(x)

Oh... I'm kinda wondering what you put for Var(y)

Since x and u are not independant

var(x+u)=var(x)+var(u)+2*cov(x,u)

 

[Nebor]

Some people have Vietnam flashbacks, I'm too young for that. I have statistics flash backs. My classmates were falling around me, they couldn't stay awake to study, they were dropping like flys! I lost half my class to that exam. I... I still can't go to gallup.com, it's too intense.

 

[TuxDave]

Originally posted by: Nebor
Some people have Vietnam flashbacks, I'm too young for that. I have statistics flash backs. My classmates were falling around me, they couldn't stay awake to study, they were dropping like flys! I lost half my class to that exam. I... I still can't go to gallup.com, it's too intense.

lol.... I hate statistics too. Why can't it leave my brain already.....

 

[UncleWai]

Thanks, Cov(a,b+c) = Cov(a,b)+Cov(a,c) that's what I need
I've looked from my stat and my econometric book, and the class notes, I can't find that identity.

var(x)+var(u)+2*cov(x,u) <--- that's what I put.

 

[TuxDave]

Originally posted by: UncleWai
Thanks, Cov(a,b+c) = Cov(a,b)+Cov(a,c) that's what I need
I've looked from my stat and my econometric book, and the class notes, I can't find that identity.

var(x)+var(u)+2*cov(x,u) <--- that's what I put.

Yeah... it's the first time I saw that identity too.

http://mathworld.wolfram.com/Covariance.html

Line (8)

Whatever... haha.. whatever gets the job done!

 

[UncleWai]

TuxDave is the Stat Wiz for ATOT.

 

[TuxDave]

Originally posted by: UncleWai
TuxDave is the Stat Wiz for ATOT.

:Q What a scary thought....

 

[chuckywang]

Dammit, I help out with stats and prob.ability too!

 

[TuxDave]

Originally posted by: chuckywang
Dammit, I help out with stats and prob.ability too!

*gives title to you*