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Solution of exercise 5

Solution of Q2:

Q3. X1 -X2v1 and X2 –X1V2 and independent

a) NZ(t) = NX1+X2 (t)= NX1 (t) NX2 (t)

(1-2t)-v1/2 (1-2t)-v2

(1-2t) –(v1+v2)/2

b) Z- X2V1+V2

Q5. X – X2V=5

0.05

0

∞

X22

P( X2V > X22) = 0.5 = 1-P(X2V<X22)=0.5

= P( X2V < X22) = 0.95 = (X2V0.95=X22)=11.0

c) 0.05/2 =0.025

Since the distribution of X21 # X22

0.025 0.025

∞

0 X22 X22

*P( X2V < X21) = 0.025 = X2V0.025=X21)=0.831

= P( X2V > X22) = 0.25

=1-P( X2V < X22) = 0.025 = P (X2V< X22)=0.975

=X2V,0.95=X22 =12.83

0.1

c) P( X2V < X21) = 0.1 = (X2V,0.1=X21)=1.61

0 x21 ∞

d)

0.01

0 ∞

X22

P( X2V > X22) = 0.1

= 1-P(X2V < X22)=0.01

P( X2V < X22) = 0.99

= X2V,0.99 = X22) = 15.09

X =TV=9

P( TV> t2) = 0.05 = 1-P( TV< t2) = 0.5

0.05

-∞ t2 ∞

P( TV< t2) = 0.95 = tv= t2 = 1.833

b)0.05/2

=0.025

Since the distribution is symmetry i.e t1=- t2

0.025 0.025

-∞ t1 t2 ∞

=-t2

P( TV> t2) = 0.025 = 1-P( TV< t2) = 0.025

P( TV< t2) = 0.975

=tv, 0.975 =t2 = 2.262

Therefor t1 = -t2= -2.262

c) 1-0.99=0.01=0.01/2

=0.005 Since the distribution is symmetry i.e t1=- t2

0.99

0.005 0.005

-∞ t1 t2 ∞

P( TV> t2) = 0.005 = 1-P( TV< t2) = 0.005

P( TV< t2)=0.995

=tv, 0.975 =t2 = 3.250

Therefor t1 = -t2= -3.250

d)

0.01

-∞ t1 ∞

P( TV<t2) = 0.01 = t1= tv,0.01) = -tv, 0.99

=-2.821

From the symmetry of T Distribution

e)

P( TV<t2) = 0.9 = t2= tv,0.9) = 1.383

0.9

t2

Q7)

X =exp(θ=ln(3)

f(x)= θe- θx =ln(3)e-ln(3)x

F(x)= 1-e- θx = 1-e-ln(3)x ; x>0

P(2≤x≤4) pdf

cdf ==e-ln(3)xdx=0.0988

=p(X ≤)= P(X≤2)=1-e-ln(3)-4 –(1-e-ln(3)-2)= 0.0988

Q8) P(X>2)

pdf

cdf ==e-Xdx= e-x/-1l2∞=

=1-p(x<2)=1-(1-e-2)=e-2

-(e∞-e-2)= e-2

Q9) X =exp (θ) = F (x)= θe- θx, x>0; F(x)=1-e- θx, x>0; E(x)= 1/ θ

P(x<E(X))=P(x< 1/ θ)=F(1/ θ)=1-e- θ (1/ θ)= 1-e-1

Q10) X= x2v = Nx(t)= (-/1-2t)v/2, t<1/2

Therefore v/2 = 1 =v=2

Or X=exp(θ) = Nx(t)= θ/ θ-t, t< θ

Therefore

1/ 1-2t=(1/2) /((1/2)-t)= θ=1/2

b) X=N( N,ð2) =NX(t)= eNT+1/2 ð2t

=N=3 and 1/2 ð2=2= ð2=4

c) X and y are indep

Nx+y (t) (2/2-t)2 y=exp(θ=2)= Ny(t)= 2/2-t

Therefore NX+y =N(t), Ny (t)

=[(2/(2-t))]3 =Nx(t)(2/(2-t))

=Nx(t)= [(2/2-t)/ (2/2-t)]= (2/2-t)2

Therefore x= gamma (α=2, β=2)

Q11) x and y are indep

NX+y(t)= e2t-1/ 2t-t2

X =exp (θ=2) = Nx(t)= 2/2-t

NX+Y(t)= NX(t) Ny(t)

e2t-1/ 2t-t2 = 2/2-t Ny(t)

Ny(t)= (e2t-1/t(2-t)) / ((2/2-t))= e2t-1/ t(2-t)-2-t/2=e2t-1/2t

E2t-e0/t(2-0)

Therefore X=uniform (a=0, b=2)