Question

3. (PMF – 8 points) Consider a sequence of independent trials of fair coin tossing. Let X denote a random variable that indicates the number of coin tosses you tried until you get heads for the first time and let y denote a random variable that indicates the number of coin tosses you tried until you get tails for the first time. For example, X = 1 and Y = 2 if you get heads on the first try and get tails on the second try, X = 3 and Y = 1 if you get tails on the first and second tries but get heads on the third try. This experiment of sequential coin tossing ends when both heads and tails appear at least once each. (a) What is the PMF of X? (b) What is the PMF of X+Y? (c) What is the PMF of max(X,Y)? (d) What is the conditional PMF of X given X > 100?

Answer 1

Cao ako nooof tosses before one gets first head X = 1 : H | K = 2 ; X = 3 : TH TTH S c, ÞCH) = \ (TH) = P(T) P(H) = 3 =(+? PCTTH) = ™CT) Þ(1)ÞEH) = (213 xax : (tykly 1*: I 2 PON=r) : % ($) PC 7K44 ) = (+9641 =643% 3 4 ... K is een z where H denotes the event of getting a head e k denotes the event of getting a tall e ; k = 1, 2, 3, ... 0 : Otherwise bcx=K) = Y=1 : T (b) y denotes the number of tosses before one gets first tail so p(T) = 1/2 cans Y=2.: HT P(HT) = 1/2 Y=3: HHT PCHAT ) = 2 PC HHH.HT) HT ! " It times - YEK K-1 times k: 1, 2, .. So, ampak 0 Otherluose ON py (Y=k) = 1

To find the p.m.f. of 2 = max (X,Y) Let F(o) be the distribution function of Z. Then $12-3) = Fz(3) - F2(-1) = P( 253) – PC 2 53-1) = P( max(x, )<3) - PC max(x,y) 53-1) = P(X&z ,Y<3) - P(K53-1, Y EZ-1) [ooter are ind pendent variables so PC x < 1, Y5X) = PCX5%). PC Y5 x)] 3 X = P(x53). PCY53) – P(x534) PCY53-1) = [P(X= 1) + PCX =2) + -- P(x = 3)] . P (Y = 1) +PCY=2) + --- P(v=31] -TP (X=1) + PCX=2) +-- 4P(x=3-1)-{p(y=1) +PCY=2) +--PCY=3-)] = true tj[***]******* 23/1 2 G1-1

- [ Use a²_b2 = Ca-bicarb) = { "*-*]***] -=['- *] ( 2 will '* 2)] ce (4.23) 17 meter 3 Now [": Z = max(x,y) & X7,; V77180 3 = 1, .,-..] 5, plz-?)= 513=1.44 OY L / ( mov(KY)=K) = neke : K=1,2,.. . 91

PCX=X) ; 2= 100,101,102, PC X4 *>100) = P( X = 2 n x 7100) P(X7,100) P(X), 100) . So, To find PCX7,100) a PCX 7100) PCX=100) + P(x = 101) +PCX=102) + ... a vyt [1+ 2+ yht -- o) ep to 2017 99-2 , vY, L) = '; 1=100, 101, 102, .. P(x = 1) - P(X3,100) 999 So, PC K=K] x 7,100) = 299-2; a =loo, lol, 102, ...00 PLEASE UPVOTE