Final Exam { Take-Home Portion SOLUTIONS. choose. Behind one of those doors is a fabulous prize. Behind each of the other two isa

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1 MATH 477 { Section E 7/29/9 Final Exam { Take-Home Portion SOLUTIONS ( pts.). A game show host has a contestant on stage and oers her three doors to choose. Behind one of those doors is a fabulous prize. Behind each of the other two isa copy of A First Course in Probability, 4 th Edition, by Sheldon Ross. The contestant rst chooses a door. The host will then open a door that she did not choose but that has a book behind it. (There is always at least one such door.) The host then says, \Would you like to stick with the door you chose originally, orwould you like to switch to the other one?" For example, she picks door #2 and the host says, \Well, if we open door #, we see that it has no fabulous prize behind it. (But isn't that a lovely blue cover.) Would you like to stick with door #2 or would you like to switch to door #3?" Should she switch? Explain your answer. Assume that the fabulous prize is something much more desirable than the probability book. Solution. Let W be the event that the contestant wins. Let S be the event that she switches and S c the event that she doesn't switch. Suppose she doesn't switch. Then the only way she can win is to choose the prize to begin with. Because there is one prize and two books, the probability that she wins is 3. Thus IP fw j S c g 3 Now, let us suppose she switches. If she picked the prize originally, then she will switch, pick a book and lose. If she did not pick the prize originally, then she picked a door with a book behind. The host will then reveal the other door with a book behind and she will switch, choosing the remaining door, the one with the prize behind. Thus she will win only if she picked a book originally. There are two books and one prize, so IP fw j Sg 2 3 Thus, she should switch. This is often referred to as the Monty Hall Problem, named after the host of the game show Let's Make a Deal. 2. Consider the following joint cumulative distribution function. F (x; y) xy ( + (, x)(, y)) x ; y

2 (Note The number was dierent for each exam.) Compute the following (3 pts.) a.) f(x; y), the joint p.d.f., Solution. We take the mixed partial derivative to obtain the density. f(x; F (x; So, f(x; y) whenever <x< and 2 F [x ( + (, x)(, y)) + xy [x ( + (, x)(, +(, x)(, 2y)+x (,(, 2y)) +(, 2x)(, 2y) (6 pts.) b.) f (x) and f Y (y), the marginal p.d.f.s of and Y, respectively, Solution. We integrate the joint density function. If x, then f (x) f(x; y) dy ( + (, 2x)(, 2y)) dy h y + (, 2x) y, y 2 i +(, 2x)(, ) And, of course, f (x) ifx orx. Writing it out, ( ; if <x<; f (x) ; otherwise. By symmetry, f Y (y) ( ; if <y<; ; otherwise. Thus, both and Y are uniform random variables over (; ). (6 pts.) c.) f jy (x j y) and f Y j (y j x), the conditional p.d.f.s, 2

3 Solution. We just use direct computation f(x; y) f jy (x j y) f Y (y) ( +(, 2x)(, 2y); if x and <y<; ; otherwise. By symmetry, ( +(, 2x)(, 2y); if <x< and y ; f Y j (y j x) ; otherwise. (4 pts.) d.) Cov(; Y ), the covariance of and Y, Solution. IE[] IE[Y ] 2 We know, because both and Y are uniform (; ) random variables, that So, we compute IE[Y ]. Therefore, IE[Y ] "x y2 ; " 3, 6 3, (xy)(+(, 2x)(, 2y)) dydx xy + xy, 2x2 y, 2xy 2 +4x 2 + y 2 dydx 2 + y2 xy2, 2x2 2 2, 2xy3 3 2 x, 6 x + 3 x2 x # x 3 3 dx Cov(; Y )IE[Y ], IE[]IE[Y ] , 2 # y3 +4x2 dx ( pts.) e.) Var() and Var(Y ), Solution. Since both and Y are uniform (; ) random variables, Var() Var(Y ) 2 3

4 (3 pts.) f.) (; Y ), the correlation between and Y. Solution. By direct computation, (; Y ) Cov(; Y ) q Var()Var(Y ) 36 q (2)(2) 3 Note that, in order to have the density function remain nonnegative. 3. Consider the following probability mass function Compute the following (5 pts.) a.) c, IP f i; Y jg p(i; j) c, 2,i j i ; ;;N, j ; ; 2; Solution. The easiest way to compute this is Therefore, c 2 N,. N, i j c, 2,i j N, c, (, 2,i ) i N, c i 2 i c 2N, 2, c 2 N, (5 pts.) b.) p (i), the marginal probability mass function of, Solution. We compute this directly p (i) j, 2,i j 2 N, 2 N,, (, 2,i ) 2 i 2 N, 4

5 (5 pts.) c.) p Y j (j j i), the conditional probability mass function of Y with respect to i, and Solution. We compute this in the usual way. p Y j (j j i) p(i; j) p (i) (2N, ), (, 2,i ) j (2 N, ), 2 i 2,i, 2,i j (5 pts.) d.) p Y (), the marginal probability mass function of Y,evaluated at j. Solution. We compute this directly as well. p Y () N, i, 2,i 2 N, N, 2 N, i N 2 N,, 2 N, + N, 2+2,N 2 N, N,, 2 N, i 2,N, 2,, 2,i 4. Consider a sequence of n bits { numbers that are either or. These represent a binary number as follows If the sequence is (a ;a 2 ;;a n ), then the number is For example (; ; ; ; ; ; ; ; ; ) gives a 2 + a a n 2 n, We will create such anumber at random by making each bit with probability p and with probability, p. ( pts.) a.) What is the expected value of the number that we create in this fashion? 5

6 Solution. Let the i th binary digit (i n, ) be denoted i. Then i is Bernoulli with parameter p. The number obtained is P n, i i 2 i. Therefore, by linearity of expectation, " n, # IE[] IE i 2 i n, i n, i n, i i IE h 2 i i i 2 i IE [ i ] 2 i p p 2n, 2, p (2n, ) ( pts.) b.) If all the bits are determined independently, what is the variance of the number that we create in this fashion? Solution. We will use the fact that for independent random variables, the variance of a sum is the sum of the variances. So,! n, Var() Var i 2 i n, i n, i i 2 2i Var ( i ) 2 2i p(, p) p(, p) 4n, 4, 3 p(, p)(4n, ) 6

7 ( pts.) 5. Two independent measurements, and Y, are taken of a quantity. IE[] IE[Y ], but Y, where 2 Var() and 2 Y Var(Y ). The two measurements are combined by means of a weighted average to give where is a scalar and. Z +(, )Y Show that IE[Z] and nd in terms of 2 and 2 Y to minimize Var(Z). Solution. First, IE[Z] IE[ +(, )Y ]IE[]+(, )IE[Y ] +(, ) Now, we compute Var(Z). We use the independence of and Y to obtain Now we look at the derivatives of Var(Z). Var(Z) Var ( +(, )Y ) 2 Var()+(, ) 2 Var(Y ) 2 2 +(, ) 2 2 Y d d Var(Z) 22 +2(, ) 2 Y d 2 d Var(Z) Y 2 > So, if we nd the that makes d Var(Z), then that minimizes Var(Z). So we want d Y, 22 Y Y, 2 Y We plug in this value of and obtain 2 Y Y Var(Z) 2 2 Y Y 7

8 ( pts.) 6. Prove that IE[] IE[ j <a]ip f <ag + IE[ j a]ip f ag Hint Dene an appropriate random variable and then compute IE[] by conditioning on it. Solution. Let Y ( ; if <a; ; if a So, by the tower property, IE[] IE [IE[ j Y ]]. But, Y is discrete, so IE [IE[ j Y ]] y IE[ j Y y]ip fy yg IE[ j Y ]IP fy g + IE[ j Y ]IP fy g IE[ j Y ]IP f <ag + IE[ j Y ]IP f ag IE[ j <a]ip f <ag + IE[ j a]ip f ag ( pts.) 7. Let have moment generating function M(t), and dene (t) log M(t). Show that (t)j t Var() Solution. Let's compute the derivatives. (t) log M(t) (t) M (t) M(t) (t) M(t)M (t), M (t)m (t) (M(t)) 2 () M()M (), (M ()) 2 (M()) 2 IE [ 2 ], (IE[]) 2 i 2 IE h 2, (IE[]) 2 Var()

9 ( pts.). For a standard normal random variable Z let n IE[Z n ]. Show that n >< > ; when n is odd; (2j)! ; 2 j j! when n 2j. Hint Start by expanding the moment generating function of Z into a Taylor series about. Solution. For any random variable (for which the moment generating function exists on an open interval containing ), M (t) n M (n) ()t n n! n But, we know for, a standard normal random variable, M (t) e t2 2 j (t 2 2) j j! IE [ n ] t n j n! t 2j 2 j j! So, we now match coecients and IE [ n ] n! IE [ n ] >< > >< > ; if n is odd; ; 2 j j! if n 2j. ; if n is odd; (2j)! ; 2 j j! if n 2j. 9

10 ( pts.) Bonus. A die is continually rolled until the total sum of all rolls exceeds 6. Approximate the probability that at least 6 rolls are necessary. Hint Use the result of the Central Limit Theorem. Solution. If at least 6 rolls are necessary, then the total of the rst 59 rolls is at most 6. Let i be the result of roll i. We know that the i 's are independent and IE[ i ]72 and Var( i )352. So, the probability is IP ( 59 i i 6 by the Central Limit Theorem. ) IP IP IP ( 59 < < i i, (59) 7 2 6, (59)7 2 P 9 59 i i, (59)(72) 6, (59)(72) q p q p ; 9 P 59 i i, (59)(72) q p (22) 973 ) ;