Math Spring Practice for the Second Exam.

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1 Math 4 - Spring 27 - Practice for the Second Exam.. Let X be a random variable and let F X be the distribution function of X: t < t 2 t < 4 F X (t) : + t t < t < 4 t. Find P(X ), P(X ), P(X 2), P(X ), P(X 4). 2. Find P(X 2 ), P(X ), P( 4 < X < ). 2. (Honors) Let X be a continuous random variable and let F X be the CDF function of X. Let Y : e F X(X). Find the density of Y.. (Honors) Let X be a random varible. Show that F X is a right-continuous function. 4. The lifetime of a transistor in hours is a random variable having a probability density function given by f X (t) c if t > and otherwise. Assume that there are t 2 5 identical transistors operationg and that they work (or not) independent of each other. What is the probability that exactly 2 of the 5 will have to be replaced within hours of the operation? 5. Let X be an exponential random variable with expectation. Find the density of Y : X 2 and compute the probabilities P(Y + t Y t) P(X + t X t) for every t > 6. A point is chosen randomly on a line segment of lenght. Find the probability that the ratio of the longer to the shorter segment is greater than If X is a normal random variable with mean 5 and variance 4, compute. P(X 7), 2. P(X ),. P( X 5 4).

2 8. (Honors) Let X be a positive continuous random variable. The hazard rate function of X is defined as λ(t) : f X(t) F X (t). (You car read the quantity λ(t) as follows: Suppose that an item survived till time t, λ(t) dt is the probability that the item will not survive an additional time dt). Show that F X (t) e λ(t)dt. Show that if X is exponential then λ(t) t for every t >. 9. (Honors) From a distance one to the ground choose an ungle (between π 2 and π 2 ) and start a ray. Let X be the (random) point that the ray hits the ground. Find the density f X (This is known as the Cauchy distribution) and compute the expectation of X.. Let X be a random variable with f X (t) ce t, < t <. Find c >, compute the CDF function and the probability P( X 2). Let Y : X. Compute the CDF and the PDF of Y. What is the distribution Y?. Let n be an integer, a > be a real number and let Γ(a) : Use integration by parts to show that t a e t dt. Γ(n) (n )!. Let X be a random variable with PDF given by f(x) : cλe λx (λx) n if x > and otherwise. Find c and compute E[X]. 2. A man is trapped in a labyrinth and he has in front of him three options. The first one will lead him to freedom in min in expectation, the second one in min in expectation and the third one in 5 min in expectation. He choose an option in random. Compute the expected time till he will be free.. Let X be a hypergeometric random variables with parameters (N, n, m). Recall the definition: We choose n balls from a total of N balls where m are white and N m are black. (Suppose here that n < m). Let X be the number of white balls selected. Show that ( m )( N m ) l m k p X (k) : ( N, k n n) Set X i be the random vaiable that takes value if the i-th white ball is selected and otherwise. Argue that X is the sum of the X i s i m and use this to compute the expectation of X. 2

3 4. Let X, X 2 be two independent Binomial random variables with parameters n, p. Show that X + X 2 is also a binomial random variable with parameters 2n, p. Compute the conditional probability mass function of X given X + X 2 m. Compute the conditional expectaion of X given X + X 2 m. Bring a Blue Book with you. Additional office hours: Monday March 27th - am The first exam will be on Chapter 2 sections 4 till 7, Chapter, sections till and Chapter 4 section. The date of the exam is March (in class). Additional office hours Monday March 27th, - a.m. It Hints/Solutions. Since F X is continuous on, 2, 4, we have that P(X ) P(X 4) P(X 2). On X, we measure the size of the jump and we get e.g. that P(X ). So (e.g.) 4 P(X ) P(X > ) + P(X ) 4 + P(X ) 5 4 F X() Set W : F X (X). Then for all t [, ], F W (t) P(W t) P(X F X (t)) F X(F (t)) t. So W is uniform on [, ] (take derivative to confirm). Then for every s [, e], F Y (s) P(Y s) P(e X s) P(X log s) F X (log(s)) log(s). X Then f Y (s) F Y (s) s, s e.. Let x n be a decreasing sequance that converges to x. Then the events A n : {X x n } are decreasing events and X x : lim n {X x n }. Then by the continuity property of probability we get that F X (x) P(X x) P( lim n {X x n }) lim n P(X x n ) lim n F X (x n ).

4 4. Note that c t dt 2 c[ /t] c/ or c. Let E i be the event that the i-th transistor will stop working within hours. Then dt P(E i ) t 2 2. Then the probability that we want is (consider binomial with parameters (5, 2 ) ) is ( 5 2 ) ( 2 ) 2 ( ). 5. By the memoryless property of the exponential we get that We have that F X (t) e t. P(X t + X t) P(X ) F X () e. F Y (t) : P(Y t) P(X t) F X ( t) e t. So, P(Y t + Y t) P(Y t + ) P(Y t) F Y (t + ) F y (t) e t t+. 6. We have that R : X when X and R X, when X. So we have X 2 X 2 P(R 4) P(X 2 )P(R 4 X 2 ) + P(X 2 )P(R 4 X 2 ) 2 P( X X 4 X 2 ) + 2 P( X X 4 X 2 ) 2 P(X 5 X 2 ) + 2 P(X 4 5 X 2 ). We compute that P(X 5 X 2 ) P(X 5 ) P(X 2 ) 2 5 P(X 4 5 X 2 ). 7. Recall that if X is a normal random variable with mean µ and variance σ 2, then is distributed as standard normal Z. So (e.g.) X µ σ P( X 5 4) P( X 9) P( 2 Z 2) Φ(2) Φ( 2) 2φ(2). 4

5 8. We have that λ(t) ( F X) F X or λ(t) (log ( F X (t))) or s λ(t)dt log ( F X (s)) or F X (t) e s λ(t)dt. If X is exponential with parameter λ then λ(t) λe λt ( e λt ) λ. 9. Since Θ is uniform on [ π, π] we have that f 2 2 Θ(t) if t [ π, π ] and otherwise. π 2 2 So F Θ (t) π ds 2 + t π, t [ π 2, π 2 ] Then π 2 F X (x) P(X x) P(tan Θ x) P(Θ tan (x)) 2 + tan (x). π Taking derivatives we get that Also, for every a > π a f X (t) π( + x 2 ). x + x dx (set 2 x2 y) 2π which forces the expectation to be infinite.. We have that We have that if t, then If t, then, We have that F X (t) f X (t)dt 2c 2 e s ds 2 F X (t) t a 2 + y dy e t dt 2c or c 2. e s ds 2 e s ds 2 e t. ( e t ). P(Y t) P( X t) P( t X t) 2 t e s ds e s F W (t) where W is an exponential with expectation. So Y is an exponential with λ. 5

6 . We have that Γ(n) t n ( e t) dt [ t n e t ] The result follows using induction. We have that Moreover, c λγ(n) (λx) n λe λx dx c E[X] Γ(n) (λx) n λe λx dx λγ(n) +(n ) t n 2 e t dt (n )Γ(n ). u n e u du cγ(n) or c Γ(n). x(λx) n λe λx dx (u) n e u du Γ(n + ) λγ(n) n λ. 2. Let X be the time needed to get out. Let E i be the event that he choose the i-th option. By the total expectation law we have that E[X] P(E i )E[X E i ] ( + + 5).. We have that So, i E[X i ] P(X i ) E[X] m i ( N ) )( n ( N n) n N. E[X i ] mn N. 4. X measures the number of S in n independent trials and X 2 the number of succes in n (indepedent of the previous experiment and to each other) trials. So it is like we perform 2n independent trials, and X + X 2 measures the total number of success. On all the cases the chance of succes were p so X + X 2 is binomial with parameters (2n, p). Also we have that P(X k X + X 2 m) P(X k and X + X 2 m) P(X + X 2 m) P(X k and X 2 m k) P(X k)p(x 2 m k) P(X + X 2 m) P(X + X 2 m) ( n k) p k ( p) n k( ) ( n m k p m k ( p) n m+k n n ) ( 2n k)( m k ( m) pm ( p) 2n m 2n m) So, the conditional distribution of X given X + X 2 m is hypergeometric (see exercise ) with parameters (2n, k, n). We use exercise to compute the expectation. 6