n x u x v n x = (u+v) n. p x (1 p) n x x = ( e t p+(1 p) ) n., x = 0,1,2,... λe = e λ t ) x = e λ e λet = e λ(et 1).

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1 Eercise 8 We can write E(X b) 2 = E ( X EX +EX b ) 2 = E ( (X EX)+(EX b) ) 2 = E(X EX) 2 +(EX b) 2 +2E ( (X EX)(EX b) ) = E(X EX) 2 +(EX b) 2 +2(EX b)e(x EX) = E(X EX) 2 +(EX b) 2 as E(X EX) = This is minimized when(ex b) 2 =, that is,ex = b Eercise 9 var(ax +b) = E(aX +b) 2 {E(aX +b)} 2 = E(a 2 X 2 +2abX +b 2 ) (a 2 {EX} 2 +2abEX +b 2 ) = a 2 EX 2 a 2 {EX} 2 = a 2 var(x) Eercise Let X Bin(n, p) The pmf of X is ( ) n P(X = ) = p ( p) n, =,,,n To obtain the mgf ofx we will use the Bionomial formula ( ) n u v n = (u+v) n we may write = M X (t) = E ( e tx) ( ) = e t n p ( p) n = = = ( ) n (e t p ) ( p) n = ( e t p+( p) ) n Now, assume thatx Poisson(λ) Then P(X = ) = e λ λ, =,,2,! Here we will use the Taylor series epansion ofe u at zero, ie, e u u =! we have = M X (t) = E ( e tx) = e te λ λ! = ( λe = e λ t )! = = e λ e λet = e λ(et ) 6

2 where S is a random variable denoting smoking status such that S(s) =, S(q) = 2, S(n) = 3 and G is a random variable denoting gender such that G(m) =, G(f) = In the table we have the distribution ofx = (G,S) as well as marginal distributions ofgand ofs p G () = 6 2 p X (,) = 2 3 p S ()+p S (2) = 3+37 = 67 4 p X (,)+p X (,2) = +5 = 5 Eercise 4 Here we have conditional probabilities p S (S = G = ) = 2 6 = 3 2 p G (G = S = ) = 3 = 3 Eercise 5 The first equality is a special case of Lemma 2 withg(y) = Y To show the second equality we will work on the RHS of the equation var(y X) = E(Y 2 X) [E(Y X)] 2 So Also, E[var(Y X)] = E[E(Y 2 X)] E{[E(Y X)] 2 } = E(Y 2 ) E{[E(Y X)] 2 } var(e[y X]) = E{[E(Y X)] 2 } {E[E(Y X)]} 2 = E{[E(Y X)] 2 } [E(Y)] 2 adding these two epressions we obtain E[var(Y X)]+var(E[Y X]) = E(Y 2 ) [E(Y)] 2 = var(y) Eercise 6 (See eample 25) Denote by a success a message which gets into the server and by X the number of successes in Y trials Then X Y Bin(Y,p), p = 3 As in Eample 25 Hence Y Poisson(λ) λ = 2 X Poisson(λp) = 7 E(X) = λp, var(x) = λp = 7 Eercise 7 Here we have a two-dimensional rv X = (X,X 2 ) which denotes length of life of two components, with joint pdf equal to f X (, 2 ) = 8 e 2 ( + 2 ) I {(, 2 ): >, 2 >} 8

3 Probability that the length of life of each of the two components will be greater than hours is: P X (X >,X 2 > ) = 8 e 2 ( + 2 ) d d 2 = 8 = 8 e 2 2 = 3 2 e e 2 d d 2 }{{} =6e /2 (byparts) e 2 2 ( 6e 2) d2 That is, the probability that the length of life of each of the two components is greater than hours is 3 2 e 2 Now we are interested in component II only, so we need to calculate the marginal pdf of the length of its life f X2 ( 2 ) = 8 e 2 ( + 2 ) d = 8 e 2 2 e 2 d }{{} = 2 e 2 2 =4 (by parts) Now, P X2 (X 2 > 2) = 2 2 e 2 2 d 2 = e That is, the probability that the length of life of component II will be bigger than 2 hours is e 3 We can write f X (, 2 ) = ( ) ( ) 8 e 2 e 2 2 = g( )h( 2 ) The joint pdf can be written as a product of two functions, one depending on only and the other on 2 only for all pairs ( 2 ) and the domains are independent the random variablesx and X 2 are independent 4 Writeg(X ) = X and h(x 2 ) = X 2 Then, by part 2 of Theorem 3 we have [ ] E[g(X )h(x 2 )] = E[g(X )]E[h(X 2 )] = E E[X 2 ] The marginal pdfs forx and X 2, respectively, are X f X ( ) = 4 e 2, f X2 ( 2 ) = 2 e 2 2 Finally, [ ] E = e 2 d X 4 E[X 2 ] = 2 E [ X2 X ] = 4 e 2 d = 2 2 e 2 2 d 2 = 2e [ ] = E E[X 2 ] = e X 9

4 Eercise 8 Take the nonnegative function oft, g(t) = var(tx +X 2 ) Then, g(t) = var(tx +X 2 ) = E [ (tx +X 2 (tµ +µ 2 )) 2] = E [ (t(x µ )+(X 2 µ 2 )) 2] = E [ ((X µ ) 2 t 2 +2(X µ )(X 2 µ 2 )t+(x 2 µ 2 ) 2] = var(x ) t 2 +2cov(X,X 2 ) t+var(x 2 ) g(t), hence there is one or no real roots of this quadratic function The discriminant is: that is and so = 4cov 2 (X,X 2 ) 4var(X )var(x 2 ) cov 2 (X,X 2 ) var(x )var(x 2 ), cov2 (X,X 2 ) var(x )var(x 2 ) ρ(x,x 2 ) 2 ρ(x,x 2 ) = if and only if the discriminant is equal to zero That is ρ(x,x 2 ) = if and only if g(t) has a single root But since (t(x µ ) + (X 2 µ 2 )) 2, the epected value g(t) = E [ (t(x µ )+(X 2 µ 2 )) 2] = if and only if This is equivalent to P ( [t(x µ )+(X 2 µ 2 )] 2 = ) = P (t(x µ )+(X 2 µ 2 ) = ) = It means thatp(x 2 = ax +b) = witha = t andb = µ t+µ 2, wheretis the root ofg(t), that is t = cov(x,x 2 ) var(x ) a = t has the same sign asρ(x,x 2 ) Eercise 9 We have { 8y, for < y ; f X,Y (,y) =, otherwise (a) VariablesX and Y are not independent because their ranges are not independent (b) We have cov(x,y) = E(XY) E(X)E(Y) E(XY) = y8ydyd = 4 9 To calculatee(x) and E(Y) we need the marginal pdfs forx and for Y f X () = and f Y (y) = 8ydy = 4( 2 ) on (,); Now we can calculate the epectations, that is, E(X) = 8yd = 4y 3 on (,) 4( 2 )d = 8 5, E(Y) = y4y 3 dy = 4 5 cov(x,y) =

5 (c) The transformation and the inverses are, respectively, u = y and v = y; and Then, Jacobian of the transformation is = uy = uv and y = v J = det ( v u ) = v So, by the theorem for a bivariate transformation we can write The support of(u,v)is f U,V (u,v) = f X,Y (uv,v) J = 8uv 3 D = {(u,v) : u <, < v } (d) The joint pdf for (U,V) is a product of two functions: one depends on u only and the other on v only Also, the ranges of U and V are independent U and V are independent random variables (e) U and V are independent, hence cov(u, V) = Eercise 2 (a) Hereu = +y and v = +y the inverses are = uv and y = v uv The Jacobian of the transformation isj = v Furthermore, the joint pdf of(x,y) (by independence) is by the transformation theorem we get and the support for(u,v) is f X,Y (,y) = λ 2 e λ(+y) f U,V (u,v) = λ 2 e λv v D = {(u,v) : < u <, < v < } (b) Random variables U and V are independent because their joint pdf can be written as a product of two functions, one depending on v only and the other on u only (this is a constant) and their domains are independent (c) We will find the marginal pdf foru f U (u) = on(, ) That is U Uniform(, ) λ 2 e λv vdv = (d) A sum of two identically distributed eponential random variables has the Erlang(2, λ) distribution