Probability, Discrete Random Variables question confusion.

Hello all,

need some help with this question.

It goes,

Suppose X is a random variable with mean "mu" and standard deviation "sigma". Consider the random variable Z = (X-mu)/sigma. Find the expected value of Z?

I put 0 but I know this is wrong.

The next part of the question wants me to find the Variance of Z. Have no idea what to do for this.

Anyone who can help with this will be greatly appreciated.

0 is correct - can you see why? How did you get to it?

For the next part, you may wish to start off by saying Var(Z) = $Var(\frac{x}{sigma} - \frac{mu}{sigma})$ amd then applying properties of variance.

Is it because E[X] = mu

mu - mu = 0

0/sigma = 0 right??

Okay I'll try and work through the 2nd part and tell you what I come up with.

My confidence in my abilities in Probability and Statistics is really shaky lol, so thank you for helping me with this.

What you've done is correct, but if I were to be pernickety you haven't laid it out properly so you would get some of the marks

E(Z) = E((x-mu)/sigma)
= E(1/sigma) * E(X-mu) = (E(X) - E(mu)) * E(1/sigma) = (mu - mu) * 1/E(sigma) using linearity and that E(X) = mu, and E(constant) = constant, so E(mu) = mu and E(sigma) = sigma, hence E(Z) = (mu-mu)/sigma = 0/sigma = 0.

Is the Variance = 1 for the 2nd part??

Hello all,

need some help with this question.

It goes,

Suppose X is a random variable with mean "mu" and standard deviation "sigma". Consider the random variable Z = (X-mu)/sigma. Find the expected value of Z?

I put 0 but I know this is wrong.

The next part of the question wants me to find the Variance of Z. Have no idea what to do for this.

Anyone who can help with this will be greatly appreciated.[/QUOTE
]you are correct, Z is normally distributed mean mu,variance=sigma^2. So N(0,1^2 )=N(0,1)
Silly question anyway,who set it?

Correct, well done

So how did you get there, and why?

OMG =O

So what I did.

Var(x/Sig - mu / Sig) = Var(x/Sig)-Var(mu/Sig)

Variance of a constant is 0.

So Var(x/Sig)-0

Then we take out (1/Sig)*Var(x)

Sig = Root of Var (x)

So Var(x)/Var(x) = 1

May have fluked it but that's what I did. Is the correct way of thinking about it.

Ahh. You're along the right lines

Remember the different properties of expectation and variance.

E(aX + bY) = aE(X) + bE(Y), and E(aX +c) = aE(X) + E(c) = aE(X) + c, as E(constant = constant).

It is not necessarily true that Var(X+Y) = Var(X) + Var(Y) as you would do for E(X+Y). But in this question we can use another property.

Side note:
The property is that we can also show that Var(aX+c) = a^2Var(X). (The constant disappears, and the scale factor is taken out as a square).

In your answer you did take out the sigma but you didn't use that Var(aX) = a^2Var(X), where a can even be a fraction.

Hopefully you have seen all of those things before. If you're unsure, ask. Otherwise you have all the tools you need to try to answer again.

Thank you so much for taking the time to help me.

You're amazing

No problem . The important thing is though, can you now see how to get to Var(Z) = 1?

My lecturer for Probability and Statistics lol.

He has given us a lot of poor questions.

I think so,

Using the properties of variance

Var(Z) = Var(aX+b) Let a = 1/Sig, b=-mu/Sig, b being a constant Var = 0

=a^2Var(x)

Sig = Root Var (x)
(1/Sig)^2 = 1/Var(x)

Var(z)=Var(x)/Var(x) = 1

you are correct, Z is normally distributed mean mu,variance=sigma^2. So N(0,1^2 )=N(0,1)
Silly question anyway,who set it?