Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




Not registered yet?

  • Index
  •  » Help Me !
  •  » Can anyone help me with this question ( urgent)?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno

Go back

Topic review (newest first)

2006-03-28 23:43:17


It's a useful result because it means that if you have such new data and you wish to recalculate the variance, you don't have to do all of the larborious summations again. You simply multiply your old result for variance by ac. Imagine you had to do this many many times on many sets of data. You'd use a computer program, but you'd be foolish to make that program recalculate all of the summations each time. You'd simply multiply each time by the new value of ac and thus save many processor ticks of work.

Further, it shows that the offset of the data by b and d has no affect on the variance. Only the multiplying factors affect the variance. Imagine the bell shaped curve, it has a certain "width" or variance. If you translate it along the x-axis left or right, this variance will not change. When you multiply each datum though, you will squash or stretch the curve depending on whether the multiplying factor is less or greater than one respectively.

2006-03-28 21:55:25

hii all...I need to answer this question urgently.

The covariance can be expressed in the following formula:-
Cov(x,y) = ∑ (x-E(x)) (y-E(y)

If we have two new variables. U=aX + b and V = cY+d where a, b, c and d are arbitrary constants.
then Cov (x,y) = ac Cov(x,y) ,    the proof of this is below.

Cov(aX + b, cY + d) = E([aX + b − E(aX + b)][cY + d − E(cY + d)])
                              = E([aX + b − aE(X) − b][cY + d − cE(Y ) − d])
                              = E(a[X − E(X)]c[Y − E(Y )])
                              = acE([X − E(X)][Y − E(Y )])
                              = acCov(X, Y )

the Question is what does the above result  (Cov (x,y) = ac Cov(x,y) )  mean?

I hope that someone here can tell me the answer.

Board footer

Powered by FluxBB