Probability, Discrete Random Variables question confusion.Watch

#1
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
3 years ago
#2
(Original post by Mr XcX)
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) = amd then applying properties of variance.
0
#3
(Original post by SeanFM)
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) = 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.
1
3 years ago
#4
(Original post by Mr XcX)
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.
0
3 years ago
#5
[QUOTE=Mr XcX;60762753]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?
1
#6
(Original post by SeanFM)
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??
0
3 years ago
#7
(Original post by Mr XcX)
Is the Variance = 1 for the 2nd part??
Correct, well done

So how did you get there, and why?
0
#8
[QUOTE=glenis72;60763125]
(Original post by Mr XcX)
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?
My lecturer for Probability and Statistics lol.

He has given us a lot of poor questions.
0
#9
(Original post by SeanFM)
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)^2*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.
0
3 years ago
#10
(Original post by Mr XcX)
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:
Spoiler:
Show
Similar to E(constant) = constant, Var(constant) = 0.

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.
0
#11
(Original post by SeanFM)
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:
Spoiler:
Show
Similar to E(constant) = constant, Var(constant) = 0.

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
0
3 years ago
#12
(Original post by Mr XcX)
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?
0
#13
(Original post by SeanFM)
No problem . The important thing is though, can you now see how to get to Var(Z) = 1?
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
0
3 years ago
#14
[QUOTE=Mr XcX;60763385]
(Original post by glenis72)

My lecturer for Probability and Statistics lol.

He has given us a lot of poor questions.
Remember Var(aX)=a^2Var(X). So Var(-X)=-1^2Var(x)=+Var(X). Var(constant)=0
So, Var(X-mu/sigma) =1/sigma^2 xVar(X-mu) = 1/sigma^2x (Var(X)+Var(mu) )=Var(X)/sigma^2=1/1=1
Remember var(X-Y)=varX+varY. Because var(X-Y)=var(X+-Y)=varX+-1^2varY=varX+varY
Not easy to explain on this Ipad
0
3 years ago
#15
(Original post by Mr XcX)
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
Perfect, well done, you seem to understand it now, as long as you remember those properties.

Again, being pernickety (and it's my fault this time) the c doesn't disappear just because Var(c) = 0, but it dissapears when calculating Var(aX+c). You could try to prove this yourself or follow a proof online, as you just need to know the properties and that Var(X) = E(X^2) - (E(X))^2.

So yes, well done, and hopefully you are feeling a bit more confident on dealing with expectation and variance.
0
3 years ago
#16
(Original post by glenis72)
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?
Not quite: mean and variance are fine but there's nothing to say Z is normally distributed.
0
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