Properties of Expected values and Variance
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1 Properties of Expected values and Variance Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011
2 Expected value Consider a random variable Y = r(x ) for some function r, e.g. Y = X so in this case r(x) = x
3 Expected value Consider a random variable Y = r(x ) for some function r, e.g. Y = X so in this case r(x) = x It turns out (and we have already used) that E(r(X )) = r(x)f (x)dx.
4 Expected value Consider a random variable Y = r(x ) for some function r, e.g. Y = X so in this case r(x) = x It turns out (and we have already used) that E(r(X )) = r(x)f (x)dx. This is not obvious since by definition E(r(X )) = xf Y (x)dx where f Y (x) is the probability density function of Y = r(x ).
5 Expected value Consider a random variable Y = r(x ) for some function r, e.g. Y = X so in this case r(x) = x It turns out (and we have already used) that E(r(X )) = r(x)f (x)dx. This is not obvious since by definition E(r(X )) = xf Y (x)dx where f Y (x) is the probability density function of Y = r(x ). You get from one integral to the other by careful uses of u substitution.
6 Expected value Consider a random variable Y = r(x ) for some function r, e.g. Y = X so in this case r(x) = x It turns out (and we have already used) that E(r(X )) = r(x)f (x)dx. This is not obvious since by definition E(r(X )) = xf Y (x)dx where f Y (x) is the probability density function of Y = r(x ). You get from one integral to the other by careful uses of u substitution. One consequence is E(aX + b) = (ax + b)f (x)dx = ae(x ) + b.
7 Expected value Consider a random variable Y = r(x ) for some function r, e.g. Y = X so in this case r(x) = x It turns out (and we have already used) that E(r(X )) = r(x)f (x)dx. This is not obvious since by definition E(r(X )) = xf Y (x)dx where f Y (x) is the probability density function of Y = r(x ). You get from one integral to the other by careful uses of u substitution. 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 )).)
8 Expected value Consider a random variable Y = r(x ) for some function r, e.g. Y = X so in this case r(x) = x It turns out (and we have already used) that E(r(X )) = r(x)f (x)dx. This is not obvious since by definition E(r(X )) = xf Y (x)dx where f Y (x) is the probability density function of Y = r(x ). You get from one integral to the other by careful uses of u substitution. 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.
9 Problem For X the uniform distribution on [0, 2] what is E(X 2 )?
10 Problem For X the uniform distribution on [0, 2] what is E(X 2 )? If X 1, X 2, X 3,...X n are random variables and Y = r(x 1, X 2, X 3,...X n ) then E(Y ) =... r(x 1, x 2, x 3,..., x n )f (x 1, x 2, x 3,..., x n )dx 1 dx 2 dx 3...dx n where f (x 1, x 2, x 3,..., x n ) is the joint probability density function.
11 Problem For X the uniform distribution on [0, 2] what is E(X 2 )? If X 1, X 2, X 3,...X n are random variables and Y = r(x 1, X 2, X 3,...X n ) then E(Y ) =... r(x 1, x 2, x 3,..., x n )f (x 1, x 2, x 3,..., x n )dx 1 dx 2 dx 3...dx n where f (x 1, x 2, x 3,..., x n ) is the joint probability density function. Problem Consider again our example of randomly choosing a point in [0, 1] [0, 1].
12 Problem For X the uniform distribution on [0, 2] what is E(X 2 )? If X 1, X 2, X 3,...X n are random variables and Y = r(x 1, X 2, X 3,...X n ) then E(Y ) =... r(x 1, x 2, x 3,..., x n )f (x 1, x 2, x 3,..., x n )dx 1 dx 2 dx 3...dx n where f (x 1, x 2, x 3,..., x n ) is the joint probability density function. Problem Consider again our example of randomly choosing a point in [0, 1] [0, 1]. We could let X be the random variable of choosing the first coordinate and Y the second. What is E(X + Y )? (note that f (x, y) = 1.)
13 Problem For X the uniform distribution on [0, 2] what is E(X 2 )? If X 1, X 2, X 3,...X n are random variables and Y = r(x 1, X 2, X 3,...X n ) then E(Y ) =... r(x 1, x 2, x 3,..., x n )f (x 1, x 2, x 3,..., x n )dx 1 dx 2 dx 3...dx n where f (x 1, x 2, x 3,..., x n ) is the joint probability density function. Problem Consider again our example of randomly choosing a point in [0, 1] [0, 1]. We could let X be the random variable of choosing the 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.
14 Properties of E(X ) A little more surprising (but not hard and we have already used): E(X 1 + X 2 + X X n ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ).
15 Properties of E(X ) A little more surprising (but not hard and we have already used): E(X 1 + X 2 + X X n ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ). Another way to look at binomial random variables; Let X i be 1 if the i th trial is a success and 0 if a failure.
16 Properties of E(X ) A little more surprising (but not hard and we have already used): E(X 1 + X 2 + X X n ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ). Another way to look at binomial random variables; Let X i be 1 if the i th trial is a success and 0 if a failure. Note that E(X i ) = 0 q + 1 p = p.
17 Properties of E(X ) A little more surprising (but not hard and we have already used): E(X 1 + X 2 + X X n ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ). Another way to look at binomial random variables; Let X i be 1 if the i th trial is a success and 0 if a failure. Note that E(X i ) = 0 q + 1 p = p. Our binomial variable (the number of successes) is X = X 1 + X 2 + X X n so E(X ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ) = np.
18 Properties of E(X ) A little more surprising (but not hard and we have already used): E(X 1 + X 2 + X X n ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ). Another way to look at binomial random variables; Let X i be 1 if the i th trial is a success and 0 if a failure. Note that E(X i ) = 0 q + 1 p = p. Our binomial variable (the number of successes) is X = X 1 + X 2 + X X n so E(X ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ) = np. What about products?
19 Properties of E(X ) A little more surprising (but not hard and we have already used): E(X 1 + X 2 + X X n ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ). Another way to look at binomial random variables; Let X i be 1 if the i th trial is a success and 0 if a failure. Note that E(X i ) = 0 q + 1 p = p. Our binomial variable (the number of successes) is X = X 1 + X 2 + X X n so E(X ) = E(X 1 ) + E(X 2 ) + E(X 3 ) E(X n ) = np. What about products? Only works out well if the random variables are independent. If X 1, X 2, X 3,...X n are independent random variables then: n n E( X i ) = E(X i ). i=1 i=1
20 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ).
21 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information!
22 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2.
23 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1.
24 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1. σ 2 (X ) = E(X 2 ) E(X ) 2 (alternative definition)
25 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1. σ 2 (X ) = E(X 2 ) E(X ) 2 (alternative definition) σ 2 (ax + b) = a 2 σ 2 (X ).
26 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1. σ 2 (X ) = E(X 2 ) E(X ) 2 (alternative definition) σ 2 (ax + b) = a 2 σ 2 (X ). Proof: σ 2 (ax + b) = E[(aX +b (aµ+b)) 2 ]
27 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1. σ 2 (X ) = E(X 2 ) E(X ) 2 (alternative definition) σ 2 (ax + b) = a 2 σ 2 (X ). Proof: σ 2 (ax + b) = E[(aX +b (aµ+b)) 2 ] = E[(aX aµ) 2 ]
28 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1. σ 2 (X ) = E(X 2 ) E(X ) 2 (alternative definition) σ 2 (ax + b) = a 2 σ 2 (X ). Proof: σ 2 (ax + b) = E[(aX +b (aµ+b)) 2 ] = E[(aX aµ) 2 ] = a 2 E[(X µ) 2 ]
29 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1. σ 2 (X ) = E(X 2 ) E(X ) 2 (alternative definition) σ 2 (ax + b) = a 2 σ 2 (X ). Proof: σ 2 (ax + b) = E[(aX +b (aµ+b)) 2 ] = E[(aX aµ) 2 ] = a 2 E[(X µ) 2 ] = a 2 σ 2 (X ).
30 Problem: Consider independent random variables X 1, X 2, and X 3, where E(X 1 ) = 2, E(X 2 ) = 1, and E(X 3 )=0. Compute E(X 2 1 (X 2 + 3X 3 ) 2 ). There is not enough information! Also assume E(X 2 1 ) = 3, E(X 2 2 ) = 1, and E(X 2 3 ) = 2. Facts about Var(X ): Var(X ) = 0 means the same as: there is a c such that Pr(X = c) = 1. σ 2 (X ) = E(X 2 ) E(X ) 2 (alternative definition) σ 2 (ax + b) = a 2 σ 2 (X ). Proof: σ 2 (ax + b) = E[(aX +b (aµ+b)) 2 ] = E[(aX aµ) 2 ] = a 2 E[(X µ) 2 ] = a 2 σ 2 (X ). For independent X 1, X 2, X 3,..., X n σ 2 (X 1 +X 2 +X X n ) = σ 2 (X 1 )+σ 2 (X 2 )+σ 2 (X 3 )+...+σ 2 (X n ).
31 Note that the last statement tells us that σ 2 (a 1 X 1 + a 2 X 2 + a 3 X a n X n ) = = a 2 1σ 2 (X 1 ) + a 2 2σ 2 (X 2 ) + a 2 3σ 2 (X 3 ) a 2 nσ 2 (X n ).
32 Note that the last statement tells us that σ 2 (a 1 X 1 + a 2 X 2 + a 3 X a n X n ) = = a 2 1σ 2 (X 1 ) + a 2 2σ 2 (X 2 ) + a 2 3σ 2 (X 3 ) a 2 nσ 2 (X n ). Now we can compute the variance of the binomial distribution with parameters n and p.
33 Note that the last statement tells us that σ 2 (a 1 X 1 + a 2 X 2 + a 3 X a n X n ) = = a 2 1σ 2 (X 1 ) + a 2 2σ 2 (X 2 ) + a 2 3σ 2 (X 3 ) a 2 nσ 2 (X n ). Now we can compute the variance of the binomial distribution with parameters n and p. As before X = X 1 + X X n where the X i are independent with Pr(X i = 0) = q and Pr(X i = 1) = p.
34 Note that the last statement tells us that σ 2 (a 1 X 1 + a 2 X 2 + a 3 X a n X n ) = = a 2 1σ 2 (X 1 ) + a 2 2σ 2 (X 2 ) + a 2 3σ 2 (X 3 ) a 2 nσ 2 (X n ). Now we can compute the variance of the binomial distribution with parameters n and p. As before X = X 1 + X X n where the X i are independent with Pr(X i = 0) = q and Pr(X i = 1) = p. So µ(x i ) = p
35 Note that the last statement tells us that σ 2 (a 1 X 1 + a 2 X 2 + a 3 X a n X n ) = = a 2 1σ 2 (X 1 ) + a 2 2σ 2 (X 2 ) + a 2 3σ 2 (X 3 ) a 2 nσ 2 (X n ). Now we can compute the variance of the binomial distribution with parameters n and p. As before X = X 1 + X X n where the X i are independent with Pr(X i = 0) = q and Pr(X i = 1) = p. So µ(x i ) = p and σ 2 (X i ) = E(X 2 i ) µ(x i ) 2 =
36 Note that the last statement tells us that σ 2 (a 1 X 1 + a 2 X 2 + a 3 X a n X n ) = = a 2 1σ 2 (X 1 ) + a 2 2σ 2 (X 2 ) + a 2 3σ 2 (X 3 ) a 2 nσ 2 (X n ). Now we can compute the variance of the binomial distribution with parameters n and p. As before X = X 1 + X X n where the X i are independent with Pr(X i = 0) = q and Pr(X i = 1) = p. So µ(x i ) = p and σ 2 (X i ) = E(X 2 i ) µ(x i ) 2 = p p 2 = p(1 p) = pq.
37 Note that the last statement tells us that σ 2 (a 1 X 1 + a 2 X 2 + a 3 X a n X n ) = = a 2 1σ 2 (X 1 ) + a 2 2σ 2 (X 2 ) + a 2 3σ 2 (X 3 ) a 2 nσ 2 (X n ). Now we can compute the variance of the binomial distribution with parameters n and p. As before X = X 1 + X X n where the X i are independent with Pr(X i = 0) = q and Pr(X i = 1) = p. So µ(x i ) = p and σ 2 (X i ) = E(Xi 2 ) µ(x i ) 2 = p p 2 = p(1 p) = pq. Thus σ 2 (X ) = Σσ 2 (X i ) = npq.
38 Consider our random variable Z which is the sum of the coordinates of a point randomly chosen from [0, 1] [0, 1].
39 Consider our random variable Z which is the sum of the coordinates of a point randomly chosen from [0, 1] [0, 1]. Z = X + Y where X and Y both represent choosing a point randomly from [0, 1].
40 Consider our random variable Z which is the sum of the coordinates of a point randomly chosen from [0, 1] [0, 1]. Z = X + Y where X and Y both represent choosing a point randomly from [0, 1]. They are independent and last time we showed σ 2 (X ) = 1 12.
41 Consider our random variable Z which is the sum of the coordinates of a point randomly chosen from [0, 1] [0, 1]. Z = X + Y where X and Y both represent choosing a point randomly from [0, 1]. They are independent and last time we showed σ 2 (X ) = So σ2 (Z) = 1 6.
42 Consider our random variable Z which is the sum of the coordinates of a point randomly chosen from [0, 1] [0, 1]. Z = X + Y where X and Y both represent choosing a point randomly from [0, 1]. They are independent and last time we showed σ 2 (X ) = So σ2 (Z) = 1 6. What is E(XY )?
43 Consider our random variable Z which is the sum of the coordinates of a point randomly chosen from [0, 1] [0, 1]. Z = X + Y where X and Y both represent choosing a point randomly from [0, 1]. They are independent and last time we showed σ 2 (X ) = So σ2 (Z) = 1 6. What is E(XY )? They are independent so E(XY ) = E(X )E(Y ) = = 1 4.
44 Consider our random variable Z which is the sum of the coordinates of a point randomly chosen from [0, 1] [0, 1]. Z = X + Y where X and Y both represent choosing a point randomly from [0, 1]. They are independent and last time we showed σ 2 (X ) = So σ2 (Z) = 1 6. What is E(XY )? They are independent so E(XY ) = E(X )E(Y ) = = 1 4. (σ 2 (XY ) is more complicated.)
45 Why is E( n i=1 X i) = n i=1 E(X i) even when not independent?
46 Why is E( n i=1 X i) = n i=1 E(X i) even when not independent? E( n X i ) = i=1... (x 1 +x x n )f (x 1, x 2,..., x n )dx 1 dx 2...dx n
47 Why is E( n i=1 X i) = n i=1 E(X i) even when not independent? E( n X i ) = i=1 =... n i=1 (x 1 +x x n )f (x 1, x 2,..., x n )dx 1 dx 2...dx n... x i f (x 1, x 2,..., x n )dx 1 dx 2...dx n.
48 Why is E( n i=1 X i) = n i=1 E(X i) even when not independent? E( n X i ) = i=1 =... n i=1 = (x 1 +x x n )f (x 1, x 2,..., x n )dx 1 dx 2...dx n... n i=1 x i f (x 1, x 2,..., x n )dx 1 dx 2...dx n. x i f i (x i )dx i = n E(X i ). i=1
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