Expectations. Expectations. (See also Hays, Appendix B; Harnett, ch. 3).


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1 Expectations Expectations. (See also Hays, Appendix B; Harnett, ch. 3). A. The expected value of a random variable is the arithmetic mean of that variable, i.e. E() = µ. As Hays notes, the idea of the expectation of a random variable began with probability theory in games of chance. Gamblers wanted to know their expected longrun winnings (or losings) if they played a game repeatedly. This term has been retained in mathematical statistics to mean the longrun average for any random variable over an indefinite number of trials or samplings. B. Discrete case: The expected value of a discrete random variable,, is found by multiplying each value by its probability and then summing over all values of the random variable. That is, if is discrete, E() = xp(x)= All µ C. Continuous case: For a continuous variable ranging over all the real numbers, the expectation is defined by E() =  xf(x) dx= µ D. Variance of : The variance of a random variable is defined as the expected (average) squared deviation of the values of this random variable about their mean. That is, V() = E[(  µ ) ] = E( ) µ = σ x In the discrete case, this is equivalent to V ( ) = σ = ( x µ ) P( x) E. Standard deviation of : The standard deviation is the positive square root of the variance, i.e. All SD ( ) = σ = σ Expectations  Page 1
2 F. Examples. 1. Hayes (p. 96) gives the probability distribution for the number of spots appearing on two fair dice. Find the mean and variance of that distribution. x p(x) xp(x) (x  µ x )5 (x  µ x )5p(x) 1/36 /36 5 5/36 3 /36 6/ /36 4 3/36 1/36 9 7/36 5 4/36 0/ /36 6 5/36 30/36 1 5/36 7 6/36 4/ /36 40/36 1 5/36 9 4/36 36/ / /36 30/36 9 7/36 11 /36 / /36 1 1/36 1/36 5 5/36 Σ xp(x) = 5/36 = 7 = µ x. The variance σ5 = 10/36 = 35/6 = 5 5/6. (NOTE: There is a simpler solution to this problem, which takes advantage of the independence of the two tosses.). Consider our earlier coin tossing experiment. If we toss a coin three times, how many times do we expect it to come up heads? And, what is the variance of this distribution? x p(x) xp(x) (x  µ x )5 (x  µ x )5p(x) 0 1/ /8 1 3/8 3/ /8 3/8 6/ /8 3 1/8 3/8.5.5/8 Σ xp(x) = 1.5. So (not surprisingly) if we toss a coin three times, we expect 1.5 heads. And, the variance = 6/8 = 3/4. Expectations  Page
3 G. EPECTATION RULES AND DEFINITIONS. a, b are any given constants., Y are random variables. The following apply. [NOTE: we ll use a few of these now and others will come in handy throughout the semester.] 1. E() = µ x = Σ xp(x) (discrete case). E(g()) = Σ g(x)p(x) = µ g() (discrete case) NOTE: g() is some function of. So, for example, if is discrete and g() =, then E( ) = Σ x p(x). 3. V() = E[(  E())5] = E(5)  E()5 = σ5 4. E(a) = a That is, the expectation of a constant is the constant, e.g. E(7) = 7 5. E(a) = a * E() e.g. if you multiple every value by, the expectation doubles. 6. E(a ± ) = a ± E() e.g. if you add 7 to every case, the expectation will increase by 7 7a. E(a ± b) = a ± be() 7b. E[(a ± ) * b] = (a ± E()) * b 8. E( + Y) = E() + E(Y). (The expectation of a sum = the sum of the expectations. This rule extends as you would expect it to when there are more than random variables, e.g. E( + Y + Z) = E() + E(Y) + E(Z)) 9. If and Y are independent, E(Y) = E()E(Y). (This rule extends as you would expect it to for more than random variables, e.g. E(YZ)=E()E(Y)E(Z).) 10. COV(,Y) = E[(  E()) * (Y  E(Y)] = E(Y)  E()E(Y) Question: What is COV(,)? 11. If and Y are independent, COV(,Y) = 0. (However, if COV(,Y) = 0, this does not necessarily mean that and Y are independent.) 1. V(a) = 0 A constant does not vary, so the variance of a constant is 0, e.g. V(7) = V(a ± ) = V() Adding a constant to a variable does not change its variance. 14. V(a ± b) = b5 * V() = σ5 b [Proof is below] 15. V( ± Y) = V() + V(Y) ± COV(,Y) = σ5 ± Y 16. If and Y are independent, V( ± Y) = V() + V(Y) However, it is generally NOT TRUE that V(Y) = V()V(Y) Expectations  Page 3
4 PROBLEMS: HINT. Keep in mind that µ and σ are constants. 1. Prove that V() = E[(  µ )5] = E(5)  µ 5. HINT: Rules 4, 5, and 8 are especially helpful here. Solution. E[(  µ ) ] = Original Formula for the variance. E(  µ + )= Expand the square µ E( E( )  E( µ )+ E( )= Rule 8: E( + Y) = E() + E(Y). That is, the expectation of a sum = Sum of the expectations µ )  µ E()+ = Rule 5: E(a) = a * E(), i.e. Expectation of a constant times a variable = The constant times the expectation of the variable; and Rule 4: E(a) = a, i.e. Expectation of a constant = the constant µ E( )  u Remember that E() = µ, hence µ E() = µ 5. QED.. Prove that V(a) = a5 * V(). HINT: Rules 3 and 5 are especially helpful. Solution. Let Y = a. Then, V(Y) = E(Y )  E(Y ) = Rule 3: V() = E[(  E())5] = E(5)  E()5 = σ5, i.e. Definition of the variance E( a ) E(a ) = Substitute for Y. Since Y = a, Y5 = a55 a E( )  a E( ) = Rule 5: E(a) = a * E(), i.e. Expectation of a constant times a variable = The constant times the expectation of the variable a (E( )  E( ) )= Factor out a5 a V() Rule 3: Definition of the variance, i.e. V() = E(5)  E()5. QED. Expectations  Page 4
5 3. Let Z = (  µ )/σ. Find E(Z) and V(Z). HINT: Apply rules 7b and 14. Solution. In this problem, a = µ, b = 1/σ. E(Z)= E  µ Definition of Z = σ E() µ Rule 7b: E[(a ± ) * b] = (a ± E()) * b. = Remember, a = µ σ, b = 1/σ. 0 Remember E() = µ, so the numerator = 0. QED Intuitively, the above makes sense; subtract the mean from every case and the new mean becomes zero. Now, for the variance, V(Z)=V  µ Definition of Z = σ 1 σ * V()= Rule 14: V(a ± b) = b5 * V() = σ5 b. Remember, b = 1/σ 1 Remember, V() = σ 5, hence σ 5 appears in both the numerator and denominator. QED. NOTE: This is called a zscore transformation. As we will see, such a transformation is extremely useful. Note that, if Z = 1, the score is one standard deviation above the mean. Expectations  Page 5
6 4. Use a different method than the one presented earlier for finding the mean and variance for the number of heads obtained in 3 coin tosses. Solution. Let 1 = 1 if the first coin toss comes up heads, 0 otherwise. In this case, 1 5 = 1 (since 15 = 1 and 05 = 0). Let and 3 be the corresponding random variables for the second and third tosses. This question is asking you to find E( ) and V( ). Note that Formula E( 1 ) = E( ) = E( 3 ) =.5 E( ) = E( 1 ) + E( ) + E( 3 ) = = 1.5. V( 1 ) = V( ) = V( 3 ) = E( 1 5)  E( 1 )5 = =.5. V( ) = V( 1 ) + V( ) + V( 3 ) = =.75 Each coin has a 50% chance of a heads Rule 8: E( + Y) = E() + E(Y), i.e. Expectation of a sum = Sum of the Expectations Rule 3: Definition of the variance. Since 1 5 = 1, E( 1 5) = E( 1 ) =.5; and E( 1 )5 =.55 =.5 Rule 16: If and Y are independent, V( ± Y) = V() + V(Y) Expectations  Page 6
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