The Student Room Group

Variance

Under the definition section of Variance:
https://en.wikipedia.org/wiki/Variance

It gives 4 lines, starting from line 1:
Var(x) = E[X-(E[X])²]....

and so fourth.

I don't understand how they get from line 2 to 3.

In line three they have:

= E[X²] - 2E[X]E[X] + (E[X]

I don't know how they got those two terms in bold. Where did the extra E[X] come from in 2E[X]E[X] (shouldn't it be E[2XE[X]]) and I don't get why (E[X] isn't E[(E[X])²]. Presumably the simplify to those terms, but it isn't explained so I need this explanation.

In the fourth line, one of the terms is cancelled. I don't get this either.

EDIT: I understand how line 4 is attained. Just don't get line 3 from line 2 now...
(edited 8 years ago)
Reply 1
Original post by djpailo
Under the definition section of Variance:
https://en.wikipedia.org/wiki/Variance

It gives 4 lines, starting from line 1:
Var(x) = E[X-(E[X])²]....

and so fourth.

I don't understand how they get from line 2 to 3.

In line three they have:

= E[X²] - 2E[X]E[X] + (E[X]

I don't know how they got those two terms in bold. Where did the extra E[X] come from in 2E[X]E[X] (shouldn't it be E[2XE[X]]) and I don't get why (E[X] isn't E[(E[X])²]. Presumably the simplify to those terms, but it isn't explained so I need this explanation.

In the fourth line, one of the terms is cancelled. I don't get this either.


I stand corrected because I am not a statistician but

E(X) = a constant

as an operator E(aX +b) = aE(X) +b

does that sort line 2 to line 3?
Reply 2
Original post by TeeEm
I stand corrected because I am not a statistician but

E(X) = a constant

as an operator E(aX +b) = aE(X) +b

does that sort line 2 to line 3?


Yes, I think you're right, thanks. I looked up that and found this link:
https://en.wikipedia.org/wiki/Expected_value#Constants

So the expected value of a constant is the constant itself. Quite literally I think this means if you take the random value to have a value of 2 in say 20 experiments, then the mean of all those twenty experiments is still 2.

I think also maybe E[E[X]] = E[X] if we say that the mean of the "mean" is simply the mean itself? If that makes sense :s. So that leaves us with what wiki gives as 2E[X]E[X]??

Is the last term then similar insofar as the the mean squared is another "constant" call a, and then the mean of that constant is just a, but of course a was the mean squared so it stays the same :s ?
Reply 3
Original post by djpailo
Yes, I think you're right, thanks. I looked up that and found this link:
https://en.wikipedia.org/wiki/Expected_value#Constants

So the expected value of a constant is the constant itself. Quite literally I think this means if you take the random value to have a value of 2 say 20 times, then the mean of all those twenty experiments is still 2.

I think also maybe E[E[X]] = E[X] if we say that the mean of the "mean" is simply the mean itself? If that makes sense :s. So that leaves us with what wiki gives as 2E[X]E[X]??

Is the last term then similar insofar as the the mean squared is another "constant" call a, and then the mean of that constant is just a, but of course a was the mean squared so it stays the same :s ?


if you replace in the wikipedia definition E(X) as µ, where µ is a constant
and note that E(aX+b) = aE(X) + b, then everything makes sense to me.
All the best

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