Split the integral into two: [0,1] and (1, )

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1 . A continuous random variable X has the iecewise df f( ) 0, 0, 0, where 0 is a ositive real number. - (a) For any real number such that 0, rove that the eected value of h( X ) X is E X. (0 ts) Solution: By definition, E[ h( X )] h( ) f ( ) d. Therefore, E X f ( ) d f ( ) d f ( ) d 0 Slit the integral into two: [0,] and (, ) d d Substitute f( ) into each 0 d d Algebra - simlify integrands 0 0 Power Rule 0 lim( b ) ( ) ( ) b F( b) F( a) [ 0] [0 ] since imlies lim( b ) 0 ( ) ( ) b ( ) ( ) Note that this = / + / when = 0. 0 Algebra The grah shows the density curve when >. The cases 0,,, and are similar in (, ), but have obvious differences in [0, ].

2 (b) Use the argument in (a) to hel confirm that f( ) is indeed a valid df. Show all work. (3 ts) Solution: First, f ( ) 0 is obvious by insection. Now, let = 0 in (a), namely f ( ) d, to obtain the identity f ( ) d. (c) Determine the median of X, and conclude that it does not deend on the value of. (Hint: Use (b) and the argument in (a). Very little calculation is necessary!) Show all work. (3 ts) Solution: The double-boed eression in (a) above,, when = 0 yields, i.e, the areas in [0, ] and (, ) are both eual to. Hence by definition, Q for any > 0. (d) Use the formula in (a) to determine the mean of X. Show all work. (Assume.) (3 ts) Solution: Let = : EX [ ]. (e) Use the formula in (a) to determine the variance of X. Show all work. (Assume,.) Algebraic simlification is not necessary. (3 ts) Solution: By definition, E X E[ X ] 4. Or, simly show that 0 f ( ) d.

3 (f) Prove that the cdf F( ) P( X ) is given by (0 ts) 0, 0, 0 F( )., Solution: Most of the integrals are just slight modifications to those in (a), with the uer limits of integration relaced by. If 0, then F( ) P( X ) f ( t) dt 0dt 0 If 0, then Thus, when =, it follows that F( ) P( X ) f ( t) dt 0 t dt t F() 0 0. Therefore, If, then F( ) P( X ) f ( t) dt t dt t ( ) - (g) Use (f) to calculate the robability that a random X falls in the interval [0.5, ], i.e., P(0.5 X ), in terms of. (3 ts) Solution: P(0.5 X ) P( X ) P( X 0.5) F() F(0.5)

4 . Two male ladybugs are sitting at oosite corners of a large window Male comosed of individual suare glass anes, as shown. At the same instant, they 0.4 start walking at the same rate along the ane edges, one edge at a time, in the hoe of eventually reaching the female sitting at the center of the window. At the start of every move along a single edge, each male randomly and indeendently selects only either the horizontal or vertical direction shown, each with the robability indicated. However, the female is only willing to wait for each male to comlete a total of n = 5 horizontal and vertical moves along the edges, before she becomes imatient and flies away. Given this information, answer each of the following Male [Hint: Let the random variable X = number of moves to the right in n = 5 indeendent moves by Male, and X = number of moves to the left in n = 5 indeendent moves by Male. ] (a) Calculate the robability of the event M = Male meets the female in n = 5 moves. Show all work. (4 ts) 5 3 P( M) P X (b) Calculate the robability of the event M = Male meets the female in n = 5 moves. Show all work. (4 ts) 5 3 P( M) P X (c) Comlete the robability chart below, including marginals. Justify all calculations. (0 ts) M M M M Part (a) yields the first column marginal robability, and likewise, art (b) yields the first row marginal robability. Moreover, via indeendence, P( M M ) P( M) P( M ) (0.3456)(0.304) = The other intersection robabilities may be similarly obtained, or more simly, by subtraction. (d) Calculate where Male is eected to be after his n = 5 moves, and indicate this osition on the diagram. Show all work. (4 ts) E[ X ] n 5(0.4) moves to the left, hence n E[ X ] n( ) 5(0.6) 3 moves uward. Therefore, combining yields the osition designated by the star symbol in the diagram. (e) With what robability will this occur? Show all work. (4 ts) 5 3 PX ( ) (0.4) (0.6)

5 In a searate scenario, the female makes n = 4 indeendent moves, each move in any direction, with the following robabilities: P (Left) 0., P (Right) 0., P (U) 0.3, P (Down) 0.4. (f) Calculate the robability that she moves in each of the four directions eactly once. Show all work. (4 ts) Multinomial distribution 4! P(Left, Right, U, Down ) (0.) (0.) (0.3) (0.4) !!!! (g) Calculate the robability that she ends u eactly where she started. Show all work. (5 ts) In addition to (f), she will also return to where she started if she moves left and right twice, or u and down twice, i.e., 4! 0 0 P(Left, Right, U 0, Down 0) (0.) (0.) (0.3) (0.4) 0.004!!0!0! 4! 0 0 P(Left 0, Right 0, U, Down ) (0.) (0.) (0.3) (0.4) !0!!! Hence, the three robabilities sum to =

6 3. (a) An eeriment involving a fair deck of 5 cards is to be conducted. Cards will be randomly selected from the deck with relacement. The random variable of interest is X = Number of trials until a face card (i.e., Jack, Queen, or King) first aears. Can the resulting outcomes be considered a seuence of Bernoulli trials? Be secific! Yes! Each outcome is indeendent of the others; since this is conducted with relacement, the robability of Success (i.e., face card ) does not change! Which robability distribution is the most aroriate to model this eeriment? Determine all arameter values. That is, X ~? Geometric distribution, with P(Success) = 5. Calculate the robability that a face card will first aear on the fourth trial. Show all work A calculator-derived answer alone will not earn any credit. The mf for this distribution is 40 (4) P( X 4) ( ) P( X ) ( ), so that (b) A second eeriment involving the same deck is to be conducted. Cards will be randomly selected from the deck, but without relacement, until a face card (i.e., Jack, Queen, or King) first aears. Can the resulting outcomes be considered a seuence of Bernoulli trials? Be secific! No! Each outcome is not indeendent of the others; since this is conducted without relacement, the robability of Success (i.e., face card ) on each trial affects that of the net trial. Calculate the robability that a face card will first aear on the fourth trial. Show all work A calculator-derived answer alone will not earn any credit Now, PX ( 4) (c) Suose that in general, a oulation of finite size N units contains s Successes. Units are to randomly drawn from the oulation without relacement. It can be formally roved that the random variable X = Number of trials until the first Success aears has mf N s ( ) P( X ) for,,3,4, N s Use this formula to recalculate the robability in (b), and show agreement in your answers. In articular, how the numerical values of the numerator and denominator. (5 ts) N = 5, s =, = 4: (4) P( X 4) , which agrees with (b)