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

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Variance of Differences of Random Variables
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[@sprencesshah]

How could I not get it in my head, I was having headache after seeing addition while subtracting variance. Finally, I can have peace of mind. Thanks

[@AloysiusPeter-d3p]

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

[@oneinabillion654]

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.

[@rohinpithwa5772]

Have you figured it out yet? My A2 stat exam is on wednesday

[@dididogster9994]

Bruh has anyone figured it out yet

[@oneinabillion654]

@@rohinpithwa5772 Sorry mate, I didn't see ur comment. Really sorry

[@oneinabillion654]

@@dididogster9994 I went through the Internet and asked many teachers, but no one could answer me. I had to dismantle everything like I'm some researcher or smth. I got the idea right before my exam. You are lucky, cuz no one will tell u

[@nicoleluo6692]

Love my TH-cam teacher 👍

[@InderjitSingh12]

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

[@shahzebafroze4093]

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

[@rebelfriend1818]

very useful thanks

[@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.

[@subramanyamujamadar1226]

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

[@jakeaustria5445]

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

[@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?

[@ЙцукенПетрович]

I think I figured the answer. Z and X are not independent, so the logic above is correct for them. For independent variables, the variance of the difference is still the sum of variances.

[@phiyahfit]

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

[@tedchirvasiu]

Ah, the good old Toaster Mic...

[@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?

[@michaelsaenz380]

Negative variance makes no sense, so your last statement is correct.

[@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

[@victor_peral]

The concept is talking about two things basically
. Expected value
. Variance ...
but it’s expected value and variance of a distribution , that distribution is a sum or difference of two distributions . z is a distribution of X+Y . The question is how do we calculated the expected value of z and it’s variance . The E(z) = E(x) + E(y) and the variance is variance of x + variance of Y.
On an occasion where random variable z = x-y the expected value of E(x)-E(y) . But the variance instead of being - is + .

[@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)))

[@Incrue]

Håkon Tjeldnes But why it is (-1)^2 ? As far as i know, (-X +Y) is the same as ((-1)(X -Y)), not ((-1)^2)(X-Y)

[@jeffstephan]

Dear Hakon, I'm stuck on the progression from E((-X+E(Y) = E(((-1^(2)(X-E(Y)). Can you you explain this for me? Much appreciated. Jeff Stephan

[@fenggeliu4241]

Which is equal to..

[@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?

[@jeppebeppebeppeson7652]

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)

[@Nina-kv4vn]

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

[@jeeth4511]

Yea

[@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).

[@rimshaaqeel1264]

plz help me about the whole chapter of probability

[@sprencesshah]

hey how are you doing after these 8 years

[@Wlooby]

Do you mean, yellow & purple??

[@pepperplume]

omg sal, i would pay to be your student

[@chopper84a]

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

[@boredgamesph4872]

Factor out the negative sign. Which is -1

[@bhoomi1318]

that was very confusing am i stupid

[@tsunningwah3471]

china

[@gracielaangulo5841]

i love the gay colors:) there awesome

[@Fongdoggie123]

shit af