S3

Original post by L'Evil Wolf

Puzzled as to why you're multiplying the Var(X) by 3.

S=Y_1+Y_2+Y_3-0.5X\\\\\text{ So }Var(S)=Var(Y_1)+Var(Y_2)+Var(Y_3)+0.25Var(X)

=Var(Y)+Var(Y)+Var(Y)+0.25Var(X)

Hence

Var(S)=3Var(Y) + 0.25Var(X)

(edited 7 years ago)

Reply 2

Original post by ghostwalker

Puzzled as to why you're multiplying the Var(X) by 3.

S=Y_1+Y_2+Y_3-0.5X\\\\\text{ So }Var(S)=Var(Y_1)+Var(Y_2)+Var(Y_3)+0.25Var(X)

Hence

Var(S)=3Var(Y) + 0.25Var(X)

How come it is not (Y1-0.5X)+(Y2-0.5X)+(Y3-0.5X)?

Reply 3

Original post by L'Evil Wolf

How come it is not (Y1-0.5X)+(Y2-0.5X)+(Y3-0.5X)?

I added an extra line to my previous post.

It's not what you've just posted because of the way S is defined. There's only one lot of X involved.

Reply 4

Original post by L'Evil Wolf

How come it is not (Y1-0.5X)+(Y2-0.5X)+(Y3-0.5X)?

The summation sign only applies to what's immediately following.

\displaystyle\sum_{i=1}^{3}Y_i - 0.5X = Y_1+Y_2+Y_3-0.5X

\displaystyle\sum_{i=1}^{3}(Y_i - 0.5X) = (Y_1-0.5X)+(Y_2-0.5X)+(Y_3-0.5X)

Reply 5

Original post by ghostwalker

I added an extra line to my previous post.

It's not what you've just posted because of the way S is defined. There's only one lot of X involved.

ah because X is not changing right?

thanks you for your help

Reply 6

Original post by L'Evil Wolf

ah because X is not changing right?

thanks you for your help

Added a second post above.

It's not because the X isn't changing.

It's really an extension of BIDMAS, or BODMAS or whatever you use. The summation is effectively a function, like log or cosine, etc.

log 6 + 7 means (log 6) + 7, not log (6+7)

Reply 7

Original post by ghostwalker

The summation sign only applies to what's immediately following.

\displaystyle\sum_{i=1}^{3}Y_i - 0.5X = Y_1+Y_2+Y_3-0.5X

\displaystyle\sum_{i=1}^{3}(Y_i - 0.5X) = (Y_1-0.5X)+(Y_2-0.5X)+(Y_3-0.5X)

Thank you very much I now understand.

Reply 8

Original post by L'Evil Wolf

Thank you very much I now understand.

Reply 9

Original post by ghostwalker

The summation sign only applies to what's immediately following.

\displaystyle\sum_{i=1}^{3}Y_i - 0.5X = Y_1+Y_2+Y_3-0.5X

\displaystyle\sum_{i=1}^{3}(Y_i - 0.5X) = (Y_1-0.5X)+(Y_2-0.5X)+(Y_3-0.5X)

I know this is a while after lol but:

S = Yi - 0.5X + Y2-0.5X + Y3 - 0.5X

= 3[Yi - 0.5X]

= 3Var(Yi)+0.25Var(X)

which still yields 28 in that regard!

So it is a conincindence that both ways that you stated result in the same answer? :smile:

Reply 10

Original post by L'Evil Wolf

How did you go from the first bolded line to the second bolded line?

Reply 11

Original post by Zacken

How did you go from the first bolded line to the second bolded line?

Took a factor of 3 out and then applied the variance of it. I assume you consider it to be wrong?

The idea stemmed from integration, how we take out constants and then integrate if that makes sense?

Reply 12

Original post by L'Evil Wolf

Took a factor of 3 out and then applied the variance of it. I assume you consider it to be wrong?

The idea stemmed from integration, how we take out constants and then integrate if that makes sense?

By that logic, why don't you "take out the 0.5 and then applied the variance of X"?

Reply 13

Original post by Zacken

By that logic, why don't you "take out the 0.5 and then applied the variance of X"?

Do you mean, have:

3[Yi - 0.5X]

3/2[2Yi - X]

3/2[ 4Var(Yi) + Var(X)]

3/2 [4(9)+4]

3/2[40}
=60

which would be incorrect.

Reply 14

Original post by L'Evil Wolf

Do you mean, have:

3[Yi - 0.5X]

3/2[2Yi - X]

3/2[ 4Var(Yi) + Var(X)]

3/2 [4(9)+4]

3/2[40}
=60

which would be incorrect.

No, I'm saying, why are you doing:

Var(3(Yi - 0.5X)) = 3Var(Yi - 0.5X).

Reply 15

Original post by Zacken

No, I'm saying, why are you doing:

Var(3(Yi - 0.5X)) = 3Var(Yi - 0.5X).

I was doing from concepts of integration, as in the factoring out a constant, but this wouldn't apply in this different mathematical operation.

Reply 16

Original post by L'Evil Wolf

I was doing from concepts of integration, as in the factoring out a constant, but this wouldn't apply in this different mathematical operation.

I'm afraid I don't understand your point, but Ghostwalker's way is the correct version and yours is simply an (incorrect) fluke.

Reply 17

Original post by Zacken

I'm afraid I don't understand your point, but Ghostwalker's way is the correct version and yours is simply an (incorrect) fluke.

Okay, thank you/

Reply 18

Original post by L'Evil Wolf

I know this is a while after lol but:

S = Yi - 0.5X + Y2-0.5X + Y3 - 0.5X

OK, you're using the second definition of S (not the one in the original question)

= 3[Yi - 0.5X]

This is meaningless; with the Yi, rather than Y1+Y2+Y3

If you're going to follow through with the second definition, you'd have:

S=Y1+Y2+y3 - 1.5X

assuming we're dealing with just one observation of X, and not three independent ones.

Working out the variance you'd have:

Var(S) = Var(Y1)+Var(Y2)+Var(Y3)+2.25Var(X)

= 3 Var(Y) + 2.25Var(X)

which won't yield 28.

IF you had three independent observations of X instead; i.e. X1,X2,X3, you'd get

S=Y1+Y2+y3 - 0.5X1-0.5X2-0.5X3

Then Var(S) = Var(Y1)+Var(Y2)+Var(Y3)+0.25Var(X1)+0.25Var(X2)+0.25Var(X3)

=3Var(Y)+0.75Var(X)

Again, no 28.

Reply 19