Variance of differences of random variables | Probability and Statistics | Khan Academy

Variance of Differences of Random Variables

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[AloysiusPeter-dp]

Thank you for making videos like these and making it available for us for free, its really helpful. Thank You Loads

[oneinabillion]

Please introduce the idea of the constants with the random variables and when to sqaure the constant and when not to. This is very important to many CIE A2 stats candidates.

[jakeaustria]

4:26 Of that fu.. Ey, sir got intrusive thoughts right there haha

[InderjitSingh]

9:41 thats all i wanted to here. thanks sal and team

[funnyvideosfans]

The prob with a lot of maths lesson is that they make use of tons of symbols completely detached from reality. You should use simpler language.

[subramanyamujamadar]

E(x) = Sum of all ( x * p(x) ).. But, what is E((x-mean)Sq) mean ?

[shahzebafroze]

thank you sir..this site has helped me a lot educational phase of my life

[phiyahfit]

How can I get personalize help on a particular problem I’m doing

[Duxa_]

Would be nice if you used a number line as part of the explanation. Im still confused a bit, I get the whole we want absolute distance, but if we got variance of 10, and we want to subtract variance of 4 from it, I dont understand why we wouldnt get variance 6. Like for instance i have an experiment which yields a result of variance of 10. The experiment consits of a variety of different parts. I find that removing part Y from the experiment reduces the variance of the experiment X by 4. So now my variance is 6 for the experiment. How would that work? Because it seems like we can only increase variance and never reduce it?

[jeppebeppebeppeson]

If the variables are dependent:

Var(X + Y) = E[(X + Y - (µx + µy))^2] = E[((X - µx) + (Y - µy))^2] = E[(X - µx)^2 + 2(X - µx)(Y - µy) + (Y - µy)^2] =
= E[(X - µx)^2] + 2E[2(X - µx)(Y - µy)] + E[(Y - µy)^2] = Var(X) + 2Cov(X, Y) + Var(Y)

Var(X - Y) = E[(X - Y - (µx - µy))^2] = E[((X - µx) - (Y - µy))^2] = E[(X - µx)^2 - 2(X - µx)(Y - µy) + (Y - µy)^2] =
= E[(X - µx)^2] - 2E[2(X - µx)(Y - µy)] + E[(Y - µy)^2] = Var(X) - 2Cov(X, Y) + Var(Y)

Note that if the variables are independent the Covariance will be 0, in which case Var(X + Y) = Var(X - Y) = Var(X) + Var(Y)

[venkataramanareddyta]

The way I think about this is that var(-y) is nothing but the difference between the squares of the -y, and the mean. Since we are squaring, -y^2 = y^2 therefore var (-y) is equal to var (y). Is my line of reasoning correct?

[ЙцукенПетрович]

I don't get it. If Z = X+ Y, then Y = Z - X. So if Var(Z) = Var(X) + Var(Y), then Var(Z-X) = Var(Z) - Var(X). Why does this logic not work? What am I missing here?

[rimshaaqeel]

plz help me about the whole chapter of probability

[Nina-kvvn]

Is this course, Probability and Statistics, available in PDF, Sal?

[Roleren]

The -1 (squared) is the same as 1, since -1*-1 = 1. He just put it there to show that you can have a minus in front of every variable(Here X and Y), and it is still the same thing.

So: E((X-E(Y))) =
 E((-X+E(Y) =
 E(((-1^(2)(X-E(Y)) =
 E((1*(X-E(Y))) =
 E((X-E(Y))) 

[choppera]

8:21 where does the minus 1 squared come from?

[phiyahfit]

This just looks like Chinese to me. I don’t understand this concept at all! Makes me wanna cry 😭. Hurts my brain wayyyy to much

[pepperplume]

omg sal, i would pay to be your student

[johnitravolta]

There's a mistake with the (-1)^2.
if you simplify this: (-1)^2 * (Y + E(-Y)^2)
you get: (-1)(-1)(Y + E(-Y)^2) = (-1)(-Y - E(-Y)^2) = (Y + E(-Y)^2).