Math 426: Probability MWF 1pm, Gasson 310 Exam 3 SOLUTIONS

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1 Name: ANSWE KEY Math 46: Probability MWF pm, Gasson Exam SOLUTIONS Problem Points Score BONUS Total 6 Please write neatly. You may leave answers below unsimplified. Have fun and write your name above!

2 . Suppose that the heights of soybean plants are normally distributed, with mean µ = 4 meters and standard deviation σ = 5/4 meters. Given a random soybean plant, use the table attached to do the following: a) Estimate the probability that the soybean plant was taller than 5 meters. b) Estimate the probability that the soybean plant was between and 4 meters. Let X indicate the height of a random soybean plant, and note that Z = X 4 5/4 is the standard normal random variable. We have P X > 5) = P Z = X 4 5/4 > 5 4 5/4 =.8), and P < X < 4) = P 4 5/4 < X 4 5/4 < 4 4 ) = P.6 < Z < ). 5/4 Each of these can be expressed in terms of the cumulative distribution function Φ of a standard normal random variable: P Z >.8) = Φ.8), and P.6 < X < ) = Φ) Φ.6) =.5 Φ.6). Each of the values Φ.8) and Φ.6) can be ascertained from the attached table, and we find Φ.8).788 and Φ.6).548. Thus P X > 5).788 =.9, and P < X < 4) =.495.

3 . Suppose that X and Y are jointly continuous with joint pdf fx, y) = x y y > x >, and zero otherwise. for a) Find the marginal density f X. For x > the marginal density is given by f X x) = fx, y) dy = x x y dy = y x x If x, then f X x) = fx, y) dy =. ) = + ) = x x x. b) Find the marginal density f Y. For y > the marginal density is given by y f Y y) = fx, y) dx = If y, then f Y y) = fx, y) dx =. x y dx = y x y ) = y ) y + = y ) y. c) Are X and Y independent? Justify your answer. For values of x, y satisfying y > x >, we have 4y ) f X x) f Y y) =. x y 4y ) The latter is equal to fx, y) only if = x y x y, i.e. xy = 4y ). This condition fails for most values of x, y satisfying y > x >, so X and Y are not independent.

4 . Suppose that X and Y are jointly continuous random variables with joint probability density function { x fx, y) = 5 y if < x < and < y < mx otherwise for some constant m >. a) Find m. Hint: The probability of the whole sample space is.) Because the probability of the whole sample space is, we have mx = P S) = fx, y) dx dy = x 5 y dy dx x 5 y mx) = m x 7 dx = dx ) = m x 8 8 = m 8 = m 6. Thus m = 4. b) Find E[XY ]. We compute: E[XY ] = xyfx, y) dx dy = ) x 6 y 4x = dx = = 4 x = 4 = 5. 4x 4 x 9 dx x 6 y dy dx 4

5 4. Prove the following or give a counterexample: If covx, Y ) =, then X and Y are independent. This statement is false. Let X be a uniform random variable on [, ], and let Y = X. Then we have Moreover, E[X] = / On the other hand, we have while E[XY ] = E[X ] = / / x dx =. /x dx =, so that covx, Y ) = E[XY ] E[X]E[Y ] =. P X > 4, Y > 6 ) = P X > 4 ) = 4 = 4, P X > 4 )P Y > 6 ) = 4 P X > 6 ) = 4 P X > 4 ) = 4 = 8. That is, P X > 4, Y > 6 ) P X > 4 ) P Y > 6 ), so X and Y are not independent. [For a discrete example, you could let X be uniform on {,, }, and Y = if X =, and Y = otherwise. Then E[XY ] = = E[X], so covx, Y ) =, but P X =, Y = ) = while P X = ) = P Y = ) = /.] 5

6 5. Suppose that X and Y are jointly continuous with joint density function { x + y if < x <, < y < fx, y) = otherwise a) Find the conditional density f X Y x y). b) Compute P X < Y = ). The conditional density f X Y x y) is given by fx,y) f Y y), so for values of y, ) we have f Y y) = fx, y) dx = x + y dx = x + xy = + y. Thus f X Y x y) = x + y + y for x, y, ), and f X Y x y) = otherwise. Since P X A Y = b) = A f X Y x b) dx for every b and A, we have P X < Y = ) / ) = f X Y x dx = = 6 5 / = 6 5 x x x 8 + ) 6 dx = ) / + = 6 5 / /) = = 7. x + dx + / ) 6

7 6. You stand on a number line at position, facing the positive direction, and play a game as follows: ) oll a fair six-sided die. ) If the result is even, you walk that many steps forward so, for example, if the result is you step forward steps) and go back to step. ) If the result is odd but not ), you walk backwards that many steps so, for example, if the result is you step backwards steps), and go back to step. 4) If the result is, you stop the game. Find the expected position when you finish. Let X indicate the position on the number line at the end of the game, and let T indicate the result of the first roll. We compute E[X] by conditioning on T : 6 E[X] = E[E[X T ]] = E[X T = k] P T = k). k= Because the die is fair, P T = k) = /6 for each k =,..., 6. Let E[X] = e. Each of the values E[X T = k] may be rewritten: E[X T = ] =, because if T = then we stop the game while standing at ; E[X T = ] = + e, because if T = then we restart the game while standing at ; similarly E[X T = ] = + e, E[X T = 4] = 4 + e, E[X T = 5] = 5 + e, and E[X T = 6] = 6 + e. Thus the above equation becomes so that e = 4. e = 6 + e) + + e) e) e) e)) = 5e + 4), 6 7

8 BONUS) Let Y be a uniform random variable on [, ], and let X be a uniform random variable on the interval [, e Y ]. That is, for a given outcome ω S, Xω) is a random number chosen uniformly from the interval [, e Y ω) ].) Find E[X]. Hint: E[X] = E[E[X Y ]].) We have E[X] = E[E[X Y ]], and the righthand side may be computed by viewing E[X Y ] as a random variable that is formed by composing the function φ : y E[X Y = y] with Y. E[X] = E[E[X Y ]] = φy)f Y y) dy = E[X Y = y]f Y y) dy = E[X Y = y] dy. Given that Y = y, X is uniform on [, e y ], and the expected value of a uniform random variable on [a, b] is the average a + b)/. That is, E[X Y = y] = +ey. This implies that + e y E[X] = dy = ) y + e y = + e + )) = e. 8