Math 426: Probability Final Exam Practice


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1 Mth 46: Probbility Finl Exm Prctice. Computtionl problems 4. Let T k (n) denote the number of prtitions of the set {,..., n} into k nonempty subsets, where k n. Argue tht T k (n) kt k (n ) + T k (n ) by considering the prtitions in which {} is subset, nd those in which is n element of subset (tht contins other elements). Solution. Every prtition of {,..., n} into k nonempty subsets deposits the element into some subset I; this subset is either singleton (i.e. I {}) or not (i.e. I \ {} ). Let us denote the prtitions in which the former occurs by I nd those prtitions in which the ltter occurs by J. Ech prtition in I consists of k nonempty subsets of the n element set {,..., n}, of which there re exctly T k (n ) possibilities. Likewise, ech of the T k (n ) prtitions of {,..., n} into k nonempty subsets determines prtition in I. Therefore, I T k (n ). On the other hnd, for ech prtition in J, we my throw out nd obtin prtition of {,..., n} into k nonempty subsets. (Note tht, fter throwing out, ech of the remining subsets is nonempty becuse we ve ssumed tht I contins elements other thn ). This defines mp from J to the set of prtitions of {,..., n} into k nonempty subsets, but for every prtition in the imge there re exctly k prtitions in J in its preimge: this is becuse to go bckwrds we need to dd to one of the k subsets in the prtition of {,..., n}. It follows tht J is equl to kt k (n ). Finlly, since every prtition of {,..., n} is in either I or J, we hve T k (n) I + J T k (n ) + kt k (n ). 5. Ten white blls nd ten blck blls re to be divided up into two urns. How cn this plcement be done so s to mximize the probbility of drwing white bll if n urn is selected t rndom, nd bll is drwn t rndom from it? Solution. Suppose we divide the blls so tht there re k blls in Urn nd k in Urn, for < k <. Such division is determined by how mny whites go in Urn ; let s cll this quntity x. Now the probbility of choosing white in the selection process is given by P (W ) P (W Urn chosen)p (Urn chosen) + P (W Urn chosen)p (Urn chosen) x k + x k ( x k + x ) k x( k) + k( x) x kx + k x kx + 5k k( k) k( k) k( k) Let us mximize P (W ) if we fix the choice of k blls in Urn. Evidently, this mens we wnt to mximize the numertor ( k)x + 5k (s function of x). If the coefficient of x is positive (i.e. k > ) this is mximized for x s lrge s possible, i.e. x k. On the other hnd, if the coefficient of x is nonpositive (i.e. k ), this is mximized when x is s smll s possible, i.e. x. In the former cse (when k > ), we get P (W ) k k + 5k k( k) 5k k k( k) 5 k k,
2 while in the ltter (when k ) we obtin P (W ) 5k k( k) 5 k. When k >, we hve 5 k > 5, so it follows tht (5 k)/( k) > 5/( k). This mens tht to mximize P (W ) we must choose k <. Now, in order to mximize P (W ) we should choose k tht mximizes (5 k)/( k). The derivtive of the ltter is ( 5 k k ) ( )( k) ( )(5 k) ( k) + 5 ( k) <, so this function of k is decresing for ll k. Tht mens tht it will be mximized when k is s smll s possible, i.e. k, in which cse P (W ) 4/9. Finlly, note tht we ve mnged to mximize P (W ) mong vlues of k in {,..., }. The finl boundry cse, in which ll the blls go in one urn, gives probbility P (W ) + /4 < 4/9, so our best scenrio is to put one white bll in one of the urns, nd the reminder in the other urn. 7. Suppose tht X tkes on the vlues,, nd. If for some constnt c we hve P (X i) cp (X i ) for i,, find E[X] (s function of c). Solution. Let P (X ) p. By ssumption, P (X ) cp, nd P (X ) cp (X ) c p. Becuse P (S), we hve P (X ) + P (X ) + P (X ) p + cp + c p p ( + c + c ), so tht p /( + c + c ). Now E[X] + P (X ) + P (X ) cp + c p p (c + c ) c + c + c + c. 9. An urn hs n white nd m blck blls. Blls re drwn, without replcement, until totl of k white blls hve been withdrwn (for k n). If X denotes the number of blls tht hve been drwn in this process, find P (X r). Solution. In order to compute P (X r) we rephrse the experiment s follows: first drw r blls, nd then drw the lst. The event tht X r is the sme s the event tht the first drw of r blls contined k white blls, nd tht the lst drw ws then white. This is then the sme s the number of wys of choosing k white blls (from n), (r ) (k ) r k blck blls (from m), nd then lst white bll (from the remining n (k )), for which there re evidently ( ) ( ) n m (n k + ) k r k possibilities. On the other hnd, the totl number of outcomes of this experiment is given by the product of the number of wys of first choosing r blls from the m+n, nd then lst one from the remining m + n (r ), so tht there re totl of ( ) m + n (m + n r + ) r
3 possible outcomes. Since ll outcomes re eqully likely, the probbility we seek is ( n )( m k P (X r) r k) (n k + ) ). (m + n r + ) ( m+n r. A continuous rndom vrible X is stndrd Cuchy if it hs density function f(x) π( + x ). Show tht if X is stndrd Cuchy then /X is stndrd Cuchy. Solution. Let Y /X. Since P (Y ) P (X /) P (X < /), the cumultive distribution function of Y is given by / F Y () F X (/) π + x dx ( tn x ) / ( ( ) tn π π ( ) π tn ( ) π tn. ( π ) ) It follows tht we cn compute the probbility density function of Y by derivting: f Y () d d (F Y ()) π + ( ) ( ) π +, so tht Y is lso stndrd Cuchy. 6. Suppose tht Z is the stndrd norml rndom vrible, I is Bernoulli rndom vrible with prmeter /3 (independent of Z), nd Y is rndom vrible given by { Z(ω) if I(ω) Y (ω) Z(ω) if I(ω) for ny outcome ω. () Are Z nd Y independent? (b) Are Y nd I independent? Solution. For prt (), let A [, ] nd B [, ]. Consider the event {Z [, ], Y [, 3]}. Since we must hve either Y (ω) Z(ω) or Y (ω) Z(ω) for every outcome ω S, the event {Z [, ], Y [, 3]} never occurs, so tht P (Z [, ], Y [, 3]). On the other hnd, consider ech of P (Z [, ]) nd P (Y [, ]). Since Z is stndrd norml, P (Z [, ]) >. By conditioning on I, we find s well P (Y [, ]) P (Y [, ] I ) P (I ) + P (Y [, ] I ) P (I ) P (Z [, ]) + P (Z [, ]) 3 3 >. This mens we hve P (Z [, ], Y [, 3]) P (Z [, ])P (Y [, 3]), so Y nd Z re not independent. 3
4 For prt (b), Y nd I re independent: if B contins both nd, then the event {Y A, I B} is the sme s the event {Y A}, nd P (Y A, I B) P (Y A) P (Y A) P (I B). If B contins neither, then evidently P (Y A, I B) P (Y A) P (I B). Now consider the event tht B contins nd not. Note tht the event {Y A, I B} is the sme s the event {Z A, I B}, since I(ω) B implies tht I(ω), so tht Y (ω) A if nd only if Z(ω) A. Thus we hve P (Y A, I B) P (Z A, I B) P (Z A) P (I B) P (Z A) 3, where we hve used the fct tht Z nd I re independent. Moreover, we my compute P (Y A) P (Y A I ) P (I ) + P (Y A I ) P (I ) P ( Z A) + P (Z A) 3 3. But becuse Z is stndrd norml, P (Z A) P ( Z A), so we conclude tht P (Y A) P (Z A). Going bck, this mens tht P (Y A, I B) P (Y A) P (I B). The sme exct considertions pply to the cse tht B contins nd not, so we conclude tht Y nd I re independent. 7. Suppose tht X nd Y re jointly continuous with joint density function c < x <, < y < x f(x, y) otherwise. for some vlue >. () Find c. (b) Find E[X]. (c) Find E[Y ]. (d) Find cov(x, Y ). Solution. For prt (), since the probbility of the whole smple spce is, we hve x x P (S) f(x, y) dx dy c dy dx c dy dx. This lst integrl is just the re of the region being integrted over, nd it s not hrd to see tht the region described by < x < nd < y < x is tringle (which is bounded by the xxis, the verticl line x, nd the line y x). This is right tringle with bse nd height, so its re is /. Since c(/), we conclude tht c /. For prts (b) nd (c), we first compute the mrginl densities: for x [, ] nd y [, ] we hve f X (x) f Y (y) f(x, y) dy f(x, y) dx x y/ dy (x ) x, nd dx Of course, f X (x) if x / [, ] nd f Y (y) if y / [, ]. 4 ( y ) ( y).
5 Now we my compute the expected vlues: E[X] xf X (x) dx x dx x3 3 3, nd y( y) E[Y ] yf Y (y) dy dy ( y y3 ) For prt (d), we hve cov(x, Y ) E[XY ] E[X]E[Y ], so we need to compute E[XY ]. x E[XY ] xyf(x, y) dx dy xy dy dx ( y x) x dx ( x x dx ) x Thus cov(x, Y ) 4 3 ( 3 4 ) /
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