properties of expected values and varianceccroke/lecture6.2.pdf · 2011. 11. 18. · properties of...
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Properties of Expected values and Variance
Christopher Croke
University of Pennsylvania
Math 115UPenn, Fall 2011
Christopher Croke Calculus 115
Expected value
Consider a random variable Y = r(X ) for some function r , e.g.Y = X 2 + 3 so in this case r(x) = x2 + 3.
It turns out (and wehave already used) that
E (r(X )) =
∫ ∞−∞
r(x)f (x)dx .
This is not obvious since by definition E (r(X )) =∫∞−∞ xfY (x)dx
where fY (x) is the probability density function of Y = r(X ).You get from one integral to the other by careful uses of usubstitution.One consequence is
E (aX + b) =
∫ ∞−∞
(ax + b)f (x)dx = aE (X ) + b.
(It is not usually the case that E (r(X )) = r(E (X )).)Similar facts old for discrete random variables.
Christopher Croke Calculus 115
Expected value
Consider a random variable Y = r(X ) for some function r , e.g.Y = X 2 + 3 so in this case r(x) = x2 + 3. It turns out (and wehave already used) that
E (r(X )) =
∫ ∞−∞
r(x)f (x)dx .
This is not obvious since by definition E (r(X )) =∫∞−∞ xfY (x)dx
where fY (x) is the probability density function of Y = r(X ).You get from one integral to the other by careful uses of usubstitution.One consequence is
E (aX + b) =
∫ ∞−∞
(ax + b)f (x)dx = aE (X ) + b.
(It is not usually the case that E (r(X )) = r(E (X )).)Similar facts old for discrete random variables.
Christopher Croke Calculus 115
Expected value
Consider a random variable Y = r(X ) for some function r , e.g.Y = X 2 + 3 so in this case r(x) = x2 + 3. It turns out (and wehave already used) that
E (r(X )) =
∫ ∞−∞
r(x)f (x)dx .
This is not obvious since by definition E (r(X )) =∫∞−∞ xfY (x)dx
where fY (x) is the probability density function of Y = r(X ).
You get from one integral to the other by careful uses of usubstitution.One consequence is
E (aX + b) =
∫ ∞−∞
(ax + b)f (x)dx = aE (X ) + b.
(It is not usually the case that E (r(X )) = r(E (X )).)Similar facts old for discrete random variables.
Christopher Croke Calculus 115
Expected value
Consider a random variable Y = r(X ) for some function r , e.g.Y = X 2 + 3 so in this case r(x) = x2 + 3. It turns out (and wehave already used) that
E (r(X )) =
∫ ∞−∞
r(x)f (x)dx .
This is not obvious since by definition E (r(X )) =∫∞−∞ xfY (x)dx
where fY (x) is the probability density function of Y = r(X ).You get from one integral to the other by careful uses of usubstitution.
One consequence is
E (aX + b) =
∫ ∞−∞
(ax + b)f (x)dx = aE (X ) + b.
(It is not usually the case that E (r(X )) = r(E (X )).)Similar facts old for discrete random variables.
Christopher Croke Calculus 115
Expected value
Consider a random variable Y = r(X ) for some function r , e.g.Y = X 2 + 3 so in this case r(x) = x2 + 3. It turns out (and wehave already used) that
E (r(X )) =
∫ ∞−∞
r(x)f (x)dx .
This is not obvious since by definition E (r(X )) =∫∞−∞ xfY (x)dx
where fY (x) is the probability density function of Y = r(X ).You get from one integral to the other by careful uses of usubstitution.One consequence is
E (aX + b) =
∫ ∞−∞
(ax + b)f (x)dx = aE (X ) + b.
(It is not usually the case that E (r(X )) = r(E (X )).)Similar facts old for discrete random variables.
Christopher Croke Calculus 115
Expected value
Consider a random variable Y = r(X ) for some function r , e.g.Y = X 2 + 3 so in this case r(x) = x2 + 3. It turns out (and wehave already used) that
E (r(X )) =
∫ ∞−∞
r(x)f (x)dx .
This is not obvious since by definition E (r(X )) =∫∞−∞ xfY (x)dx
where fY (x) is the probability density function of Y = r(X ).You get from one integral to the other by careful uses of usubstitution.One consequence is
E (aX + b) =
∫ ∞−∞
(ax + b)f (x)dx = aE (X ) + b.
(It is not usually the case that E (r(X )) = r(E (X )).)
Similar facts old for discrete random variables.
Christopher Croke Calculus 115
Expected value
Consider a random variable Y = r(X ) for some function r , e.g.Y = X 2 + 3 so in this case r(x) = x2 + 3. It turns out (and wehave already used) that
E (r(X )) =
∫ ∞−∞
r(x)f (x)dx .
This is not obvious since by definition E (r(X )) =∫∞−∞ xfY (x)dx
where fY (x) is the probability density function of Y = r(X ).You get from one integral to the other by careful uses of usubstitution.One consequence is
E (aX + b) =
∫ ∞−∞
(ax + b)f (x)dx = aE (X ) + b.
(It is not usually the case that E (r(X )) = r(E (X )).)Similar facts old for discrete random variables.
Christopher Croke Calculus 115
Problem For X the uniform distribution on [0, 2] what is E (X 2)?
If X1,X2,X3, ...Xn are random variables andY = r(X1,X2,X3, ...Xn) then
E (Y ) =
∫ ∫ ∫...
∫r(x1, x2, x3, ..., xn)f (x1, x2, x3, ..., xn)dx1dx2dx3...dxn.
where f (x1, x2, x3, ..., xn) is the joint probability density function.Problem Consider again our example of randomly choosing a pointin [0, 1]× [0, 1]. We could let X be the random variable of choosingthe first coordinate and Y the second. What is E (X + Y )?(note that f (x , y) = 1.)
Easy properties of expected values:If Pr(X ≥ a) = 1 then E (X ) ≥ a.If Pr(X ≤ b) = 1 then E (X ) ≤ b.
Christopher Croke Calculus 115
Problem For X the uniform distribution on [0, 2] what is E (X 2)?
If X1,X2,X3, ...Xn are random variables andY = r(X1,X2,X3, ...Xn) then
E (Y ) =
∫ ∫ ∫...
∫r(x1, x2, x3, ..., xn)f (x1, x2, x3, ..., xn)dx1dx2dx3...dxn.
where f (x1, x2, x3, ..., xn) is the joint probability density function.
Problem Consider again our example of randomly choosing a pointin [0, 1]× [0, 1]. We could let X be the random variable of choosingthe first coordinate and Y the second. What is E (X + Y )?(note that f (x , y) = 1.)
Easy properties of expected values:If Pr(X ≥ a) = 1 then E (X ) ≥ a.If Pr(X ≤ b) = 1 then E (X ) ≤ b.
Christopher Croke Calculus 115
Problem For X the uniform distribution on [0, 2] what is E (X 2)?
If X1,X2,X3, ...Xn are random variables andY = r(X1,X2,X3, ...Xn) then
E (Y ) =
∫ ∫ ∫...
∫r(x1, x2, x3, ..., xn)f (x1, x2, x3, ..., xn)dx1dx2dx3...dxn.
where f (x1, x2, x3, ..., xn) is the joint probability density function.Problem Consider again our example of randomly choosing a pointin [0, 1]× [0, 1].
We could let X be the random variable of choosingthe first coordinate and Y the second. What is E (X + Y )?(note that f (x , y) = 1.)
Easy properties of expected values:If Pr(X ≥ a) = 1 then E (X ) ≥ a.If Pr(X ≤ b) = 1 then E (X ) ≤ b.
Christopher Croke Calculus 115
Problem For X the uniform distribution on [0, 2] what is E (X 2)?
If X1,X2,X3, ...Xn are random variables andY = r(X1,X2,X3, ...Xn) then
E (Y ) =
∫ ∫ ∫...
∫r(x1, x2, x3, ..., xn)f (x1, x2, x3, ..., xn)dx1dx2dx3...dxn.
where f (x1, x2, x3, ..., xn) is the joint probability density function.Problem Consider again our example of randomly choosing a pointin [0, 1]× [0, 1]. We could let X be the random variable of choosingthe first coordinate and Y the second. What is E (X + Y )?(note that f (x , y) = 1.)
Easy properties of expected values:If Pr(X ≥ a) = 1 then E (X ) ≥ a.If Pr(X ≤ b) = 1 then E (X ) ≤ b.
Christopher Croke Calculus 115
Problem For X the uniform distribution on [0, 2] what is E (X 2)?
If X1,X2,X3, ...Xn are random variables andY = r(X1,X2,X3, ...Xn) then
E (Y ) =
∫ ∫ ∫...
∫r(x1, x2, x3, ..., xn)f (x1, x2, x3, ..., xn)dx1dx2dx3...dxn.
where f (x1, x2, x3, ..., xn) is the joint probability density function.Problem Consider again our example of randomly choosing a pointin [0, 1]× [0, 1]. We could let X be the random variable of choosingthe first coordinate and Y the second. What is E (X + Y )?(note that f (x , y) = 1.)
Easy properties of expected values:If Pr(X ≥ a) = 1 then E (X ) ≥ a.If Pr(X ≤ b) = 1 then E (X ) ≤ b.
Christopher Croke Calculus 115
Properties of E (X )
A little more surprising (but not hard and we have already used):
E (X1 +X2 +X3 + ...+Xn) = E (X1) +E (X2) +E (X3) + ...+E (Xn).
Another way to look at binomial random variables;Let Xi be 1 if the i th trial is a success and 0 if a failure. Note thatE (Xi ) = 0 · q + 1 · p = p.Our binomial variable (the number of successes) isX = X1 + X2 + X3 + ...+ Xn so
E (X ) = E (X1) + E (X2) + E (X3) + ...+ E (Xn) = np.
What about products? Only works out well if the random variablesare independent. If X1,X2,X3, ...Xn are independent randomvariables then:
E (n∏
i=1
Xi ) =n∏
i=1
E (Xi ).
Christopher Croke Calculus 115
Properties of E (X )
A little more surprising (but not hard and we have already used):
E (X1 +X2 +X3 + ...+Xn) = E (X1) +E (X2) +E (X3) + ...+E (Xn).
Another way to look at binomial random variables;Let Xi be 1 if the i th trial is a success and 0 if a failure.
Note thatE (Xi ) = 0 · q + 1 · p = p.Our binomial variable (the number of successes) isX = X1 + X2 + X3 + ...+ Xn so
E (X ) = E (X1) + E (X2) + E (X3) + ...+ E (Xn) = np.
What about products? Only works out well if the random variablesare independent. If X1,X2,X3, ...Xn are independent randomvariables then:
E (n∏
i=1
Xi ) =n∏
i=1
E (Xi ).
Christopher Croke Calculus 115
Properties of E (X )
A little more surprising (but not hard and we have already used):
E (X1 +X2 +X3 + ...+Xn) = E (X1) +E (X2) +E (X3) + ...+E (Xn).
Another way to look at binomial random variables;Let Xi be 1 if the i th trial is a success and 0 if a failure. Note thatE (Xi ) = 0 · q + 1 · p = p.
Our binomial variable (the number of successes) isX = X1 + X2 + X3 + ...+ Xn so
E (X ) = E (X1) + E (X2) + E (X3) + ...+ E (Xn) = np.
What about products? Only works out well if the random variablesare independent. If X1,X2,X3, ...Xn are independent randomvariables then:
E (n∏
i=1
Xi ) =n∏
i=1
E (Xi ).
Christopher Croke Calculus 115
Properties of E (X )
A little more surprising (but not hard and we have already used):
E (X1 +X2 +X3 + ...+Xn) = E (X1) +E (X2) +E (X3) + ...+E (Xn).
Another way to look at binomial random variables;Let Xi be 1 if the i th trial is a success and 0 if a failure. Note thatE (Xi ) = 0 · q + 1 · p = p.Our binomial variable (the number of successes) isX = X1 + X2 + X3 + ...+ Xn so
E (X ) = E (X1) + E (X2) + E (X3) + ...+ E (Xn) = np.
What about products? Only works out well if the random variablesare independent. If X1,X2,X3, ...Xn are independent randomvariables then:
E (n∏
i=1
Xi ) =n∏
i=1
E (Xi ).
Christopher Croke Calculus 115
Properties of E (X )
A little more surprising (but not hard and we have already used):
E (X1 +X2 +X3 + ...+Xn) = E (X1) +E (X2) +E (X3) + ...+E (Xn).
Another way to look at binomial random variables;Let Xi be 1 if the i th trial is a success and 0 if a failure. Note thatE (Xi ) = 0 · q + 1 · p = p.Our binomial variable (the number of successes) isX = X1 + X2 + X3 + ...+ Xn so
E (X ) = E (X1) + E (X2) + E (X3) + ...+ E (Xn) = np.
What about products?
Only works out well if the random variablesare independent. If X1,X2,X3, ...Xn are independent randomvariables then:
E (n∏
i=1
Xi ) =n∏
i=1
E (Xi ).
Christopher Croke Calculus 115
Properties of E (X )
A little more surprising (but not hard and we have already used):
E (X1 +X2 +X3 + ...+Xn) = E (X1) +E (X2) +E (X3) + ...+E (Xn).
Another way to look at binomial random variables;Let Xi be 1 if the i th trial is a success and 0 if a failure. Note thatE (Xi ) = 0 · q + 1 · p = p.Our binomial variable (the number of successes) isX = X1 + X2 + X3 + ...+ Xn so
E (X ) = E (X1) + E (X2) + E (X3) + ...+ E (Xn) = np.
What about products? Only works out well if the random variablesare independent. If X1,X2,X3, ...Xn are independent randomvariables then:
E (n∏
i=1
Xi ) =n∏
i=1
E (Xi ).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).
There is not enough information! Also assume E (X 21 ) = 3,
E (X 22 ) = 1, and E (X 2
3 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information!
Also assume E (X 21 ) = 3,
E (X 22 ) = 1, and E (X 2
3 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ).
Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2]
= E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2]
= a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2]
= a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Problem: Consider independent random variables X1, X2, andX3, where E (X1) = 2, E (X2) = −1, and E (X3)=0. ComputeE (X 2
1 (X2 + 3X3)2).There is not enough information! Also assume E (X 2
1 ) = 3,E (X 2
2 ) = 1, and E (X 23 ) = 2.
Facts about Var(X ):
Var(X ) = 0 means the same as: there is a c such thatPr(X = c) = 1.
σ2(X ) = E (X 2) − E (X )2 (alternative definition)
σ2(aX + b) = a2σ2(X ). Proof: σ2(aX + b) =
E [(aX+b−(aµ+b))2] = E [(aX−aµ)2] = a2E [(X−µ)2] = a2σ2(X ).
For independent X1,X2,X3, ...,Xn
σ2(X1+X2+X3+...+Xn) = σ2(X1)+σ2(X2)+σ2(X3)+...+σ2(Xn).
Christopher Croke Calculus 115
Properties of Var(X )
Note that the last statement tells us that
σ2(a1X1 + a2X2 + a3X3 + ...+ anXn) =
= a21σ2(X1) + a22σ
2(X2) + a23σ2(X3) + ...+ a2nσ
2(Xn).
Now we can compute the variance of the binomial distribution withparameters n and p. As before X = X1 + X2 + ...+ Xn where theXi are independent with Pr(Xi = 0) = q and Pr(Xi = 1) = p.So µ(Xi ) = p andσ2(Xi ) = E (X 2
i ) − µ(Xi )2 = p − p2 = p(1 − p) = pq.
Thus σ2(X ) = Σσ2(Xi ) = npq.
Christopher Croke Calculus 115
Properties of Var(X )
Note that the last statement tells us that
σ2(a1X1 + a2X2 + a3X3 + ...+ anXn) =
= a21σ2(X1) + a22σ
2(X2) + a23σ2(X3) + ...+ a2nσ
2(Xn).
Now we can compute the variance of the binomial distribution withparameters n and p.
As before X = X1 + X2 + ...+ Xn where theXi are independent with Pr(Xi = 0) = q and Pr(Xi = 1) = p.So µ(Xi ) = p andσ2(Xi ) = E (X 2
i ) − µ(Xi )2 = p − p2 = p(1 − p) = pq.
Thus σ2(X ) = Σσ2(Xi ) = npq.
Christopher Croke Calculus 115
Properties of Var(X )
Note that the last statement tells us that
σ2(a1X1 + a2X2 + a3X3 + ...+ anXn) =
= a21σ2(X1) + a22σ
2(X2) + a23σ2(X3) + ...+ a2nσ
2(Xn).
Now we can compute the variance of the binomial distribution withparameters n and p. As before X = X1 + X2 + ...+ Xn where theXi are independent with Pr(Xi = 0) = q and Pr(Xi = 1) = p.
So µ(Xi ) = p andσ2(Xi ) = E (X 2
i ) − µ(Xi )2 = p − p2 = p(1 − p) = pq.
Thus σ2(X ) = Σσ2(Xi ) = npq.
Christopher Croke Calculus 115
Properties of Var(X )
Note that the last statement tells us that
σ2(a1X1 + a2X2 + a3X3 + ...+ anXn) =
= a21σ2(X1) + a22σ
2(X2) + a23σ2(X3) + ...+ a2nσ
2(Xn).
Now we can compute the variance of the binomial distribution withparameters n and p. As before X = X1 + X2 + ...+ Xn where theXi are independent with Pr(Xi = 0) = q and Pr(Xi = 1) = p.So µ(Xi ) = p
andσ2(Xi ) = E (X 2
i ) − µ(Xi )2 = p − p2 = p(1 − p) = pq.
Thus σ2(X ) = Σσ2(Xi ) = npq.
Christopher Croke Calculus 115
Properties of Var(X )
Note that the last statement tells us that
σ2(a1X1 + a2X2 + a3X3 + ...+ anXn) =
= a21σ2(X1) + a22σ
2(X2) + a23σ2(X3) + ...+ a2nσ
2(Xn).
Now we can compute the variance of the binomial distribution withparameters n and p. As before X = X1 + X2 + ...+ Xn where theXi are independent with Pr(Xi = 0) = q and Pr(Xi = 1) = p.So µ(Xi ) = p andσ2(Xi ) = E (X 2
i ) − µ(Xi )2 =
p − p2 = p(1 − p) = pq.Thus σ2(X ) = Σσ2(Xi ) = npq.
Christopher Croke Calculus 115
Properties of Var(X )
Note that the last statement tells us that
σ2(a1X1 + a2X2 + a3X3 + ...+ anXn) =
= a21σ2(X1) + a22σ
2(X2) + a23σ2(X3) + ...+ a2nσ
2(Xn).
Now we can compute the variance of the binomial distribution withparameters n and p. As before X = X1 + X2 + ...+ Xn where theXi are independent with Pr(Xi = 0) = q and Pr(Xi = 1) = p.So µ(Xi ) = p andσ2(Xi ) = E (X 2
i ) − µ(Xi )2 = p − p2 = p(1 − p) = pq.
Thus σ2(X ) = Σσ2(Xi ) = npq.
Christopher Croke Calculus 115
Properties of Var(X )
Note that the last statement tells us that
σ2(a1X1 + a2X2 + a3X3 + ...+ anXn) =
= a21σ2(X1) + a22σ
2(X2) + a23σ2(X3) + ...+ a2nσ
2(Xn).
Now we can compute the variance of the binomial distribution withparameters n and p. As before X = X1 + X2 + ...+ Xn where theXi are independent with Pr(Xi = 0) = q and Pr(Xi = 1) = p.So µ(Xi ) = p andσ2(Xi ) = E (X 2
i ) − µ(Xi )2 = p − p2 = p(1 − p) = pq.
Thus σ2(X ) = Σσ2(Xi ) = npq.
Christopher Croke Calculus 115
Properties of Var(X )
Consider our random variable Z which is the sum of thecoordinates of a point randomly chosen from [0, 1] × [0, 1].
Z = X + Y where X and Y both represent choosing a pointrandomly from [0, 1]. They are independent and last time weshowed σ2(X ) = 1
12 . So σ2(Z ) = 16 .
What is E (XY )? They are independent soE (XY ) = E (X )E (Y ) = 1
2 · 12 = 1
4 .(σ2(XY ) is more complicated.)
Christopher Croke Calculus 115
Properties of Var(X )
Consider our random variable Z which is the sum of thecoordinates of a point randomly chosen from [0, 1] × [0, 1].Z = X + Y where X and Y both represent choosing a pointrandomly from [0, 1].
They are independent and last time weshowed σ2(X ) = 1
12 . So σ2(Z ) = 16 .
What is E (XY )? They are independent soE (XY ) = E (X )E (Y ) = 1
2 · 12 = 1
4 .(σ2(XY ) is more complicated.)
Christopher Croke Calculus 115
Properties of Var(X )
Consider our random variable Z which is the sum of thecoordinates of a point randomly chosen from [0, 1] × [0, 1].Z = X + Y where X and Y both represent choosing a pointrandomly from [0, 1]. They are independent and last time weshowed σ2(X ) = 1
12 .
So σ2(Z ) = 16 .
What is E (XY )? They are independent soE (XY ) = E (X )E (Y ) = 1
2 · 12 = 1
4 .(σ2(XY ) is more complicated.)
Christopher Croke Calculus 115
Properties of Var(X )
Consider our random variable Z which is the sum of thecoordinates of a point randomly chosen from [0, 1] × [0, 1].Z = X + Y where X and Y both represent choosing a pointrandomly from [0, 1]. They are independent and last time weshowed σ2(X ) = 1
12 . So σ2(Z ) = 16 .
What is E (XY )? They are independent soE (XY ) = E (X )E (Y ) = 1
2 · 12 = 1
4 .(σ2(XY ) is more complicated.)
Christopher Croke Calculus 115
Properties of Var(X )
Consider our random variable Z which is the sum of thecoordinates of a point randomly chosen from [0, 1] × [0, 1].Z = X + Y where X and Y both represent choosing a pointrandomly from [0, 1]. They are independent and last time weshowed σ2(X ) = 1
12 . So σ2(Z ) = 16 .
What is E (XY )?
They are independent soE (XY ) = E (X )E (Y ) = 1
2 · 12 = 1
4 .(σ2(XY ) is more complicated.)
Christopher Croke Calculus 115
Properties of Var(X )
Consider our random variable Z which is the sum of thecoordinates of a point randomly chosen from [0, 1] × [0, 1].Z = X + Y where X and Y both represent choosing a pointrandomly from [0, 1]. They are independent and last time weshowed σ2(X ) = 1
12 . So σ2(Z ) = 16 .
What is E (XY )? They are independent soE (XY ) = E (X )E (Y ) = 1
2 · 12 = 1
4 .
(σ2(XY ) is more complicated.)
Christopher Croke Calculus 115
Properties of Var(X )
Consider our random variable Z which is the sum of thecoordinates of a point randomly chosen from [0, 1] × [0, 1].Z = X + Y where X and Y both represent choosing a pointrandomly from [0, 1]. They are independent and last time weshowed σ2(X ) = 1
12 . So σ2(Z ) = 16 .
What is E (XY )? They are independent soE (XY ) = E (X )E (Y ) = 1
2 · 12 = 1
4 .(σ2(XY ) is more complicated.)
Christopher Croke Calculus 115
Why is E (∑n
i=1 Xi ) =∑n
i=1 E (Xi ) even when not independent?
E (n∑
i=1
Xi ) =
∫ ∫...
∫(x1+x2+...+xn)f (x1, x2, ..., xn)dx1dx2...dxn
=n∑
i=1
∫ ∫...
∫xi f (x1, x2, ..., xn)dx1dx2...dxn.
=n∑
i=1
∫xi fi (xi )dxi =
n∑i=1
E (Xi ).
Christopher Croke Calculus 115
Why is E (∑n
i=1 Xi ) =∑n
i=1 E (Xi ) even when not independent?
E (n∑
i=1
Xi ) =
∫ ∫...
∫(x1+x2+...+xn)f (x1, x2, ..., xn)dx1dx2...dxn
=n∑
i=1
∫ ∫...
∫xi f (x1, x2, ..., xn)dx1dx2...dxn.
=n∑
i=1
∫xi fi (xi )dxi =
n∑i=1
E (Xi ).
Christopher Croke Calculus 115
Why is E (∑n
i=1 Xi ) =∑n
i=1 E (Xi ) even when not independent?
E (n∑
i=1
Xi ) =
∫ ∫...
∫(x1+x2+...+xn)f (x1, x2, ..., xn)dx1dx2...dxn
=n∑
i=1
∫ ∫...
∫xi f (x1, x2, ..., xn)dx1dx2...dxn.
=n∑
i=1
∫xi fi (xi )dxi =
n∑i=1
E (Xi ).
Christopher Croke Calculus 115
Why is E (∑n
i=1 Xi ) =∑n
i=1 E (Xi ) even when not independent?
E (n∑
i=1
Xi ) =
∫ ∫...
∫(x1+x2+...+xn)f (x1, x2, ..., xn)dx1dx2...dxn
=n∑
i=1
∫ ∫...
∫xi f (x1, x2, ..., xn)dx1dx2...dxn.
=n∑
i=1
∫xi fi (xi )dxi =
n∑i=1
E (Xi ).
Christopher Croke Calculus 115