Var (X) = E[(X - mean)2] = E(X2) - [E(X)]2 where mean = E(X)
- Thread starter puneet_
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Expected value of X is the mean of X; they are equivalent. That being said, the Expected Value Function iteself is not the mean, for example, E(X) = the mean of X but E(aX+bX^2) would not be the mean of X, for example. It could be thought of as the mean of aX + bX^2 though!
jairamjana
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Well
Var X = E[(X- u)^2] = E[(X- E(X))^2] = E[(X^2 - 2 * X * E(X) + E(X)^2)] = E(X^2) - 2 * E(X) * E(X) + E(E(X)^2)] = E(X^2) - 2 * E(X) ^2 + E(X)^2] = E(X^2) - E(X) ^2
It's just derivation using Expectations Operator.. u is always E(X) in a probability distribution with random variable X ... summation p * f(x) will give us the mean aka E(X) ..
Var X = E[(X- u)^2] = E[(X- E(X))^2] = E[(X^2 - 2 * X * E(X) + E(X)^2)] = E(X^2) - 2 * E(X) * E(X) + E(E(X)^2)] = E(X^2) - 2 * E(X) ^2 + E(X)^2] = E(X^2) - E(X) ^2
It's just derivation using Expectations Operator.. u is always E(X) in a probability distribution with random variable X ... summation p * f(x) will give us the mean aka E(X) ..
Yes the Expected Value function is same as Mean.Let assume n different observations(x1,x2,...xn) all being equally likely to occur with probability =1/n
The observations mean is =sum of observations/n=(x1+x2+x3+.....xn)/n=(1/n)*x1+(1/n)*x2+.....+(1/n)*xn=probability of obs 1 occurance*x1+probability of obs 2 occurance*x2+.......+probability of obs n occurrence*xn=Expected value of observations(x1,x2,...xn)
therefore mean=Expected value for any set of n observations(x1,x2,...xn).
For e.g. dice throw all the faces (1,2,...6) are equally likely with probability 1/6,the mean of the faces=(1+2+3+4+5+6)/6=3.5
The expected value=probability of obs 1 occurance*1+probability of obs 2 occurance*2+.......+probability of obs 6 occurrence*6=(1/6)*1+(1/6)*2+(1/6)*3+(1/6)*4+(1/6)*5+(1/6)*6=(1+2+3+4+5+6)/6=3.5= mean of the faces.
Let check for X=1, then RHS=Var(X)=Var(1)=0
LHS=E[(X-mean)^2]=E[(1-1)^2]=E[(0)^2]=E[0]=0=RHS hence the relation Var(X)=E[(X-mean)^2] holds for X=1 thus in this way ascent to any numbers so that the equation holds for any set of numbers.
thanks
The observations mean is =sum of observations/n=(x1+x2+x3+.....xn)/n=(1/n)*x1+(1/n)*x2+.....+(1/n)*xn=probability of obs 1 occurance*x1+probability of obs 2 occurance*x2+.......+probability of obs n occurrence*xn=Expected value of observations(x1,x2,...xn)
therefore mean=Expected value for any set of n observations(x1,x2,...xn).
For e.g. dice throw all the faces (1,2,...6) are equally likely with probability 1/6,the mean of the faces=(1+2+3+4+5+6)/6=3.5
The expected value=probability of obs 1 occurance*1+probability of obs 2 occurance*2+.......+probability of obs 6 occurrence*6=(1/6)*1+(1/6)*2+(1/6)*3+(1/6)*4+(1/6)*5+(1/6)*6=(1+2+3+4+5+6)/6=3.5= mean of the faces.
Let check for X=1, then RHS=Var(X)=Var(1)=0
LHS=E[(X-mean)^2]=E[(1-1)^2]=E[(0)^2]=E[0]=0=RHS hence the relation Var(X)=E[(X-mean)^2] holds for X=1 thus in this way ascent to any numbers so that the equation holds for any set of numbers.
thanks
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