982Bfinalsol

7/24/2019 982Bfinalsol

1/5

98201-05

1. (10%) () (/) P(t) t

, P(t0) =P0 P(t) =P(t) + , >0 P(t)

Sol: tt0

P(t)

P(t) +

dt=

tt0

dt

lnP(t) +

P0+

=(t t0)

P(t) =

+ (P0+

)e(tt0)

Write down the exact solution: 10 pts.

Write down the general solution: 8 pts.

Write down or derive the formula: 5 pts.

Others : give you some points accorddng to the wrong extent.

2. (10%) dy

dt = 2ty+ 4t, y(0) = 1

Sol:

e 2tdt =et

2

(6 pts)

et2

y 2tyet2

= 4tet2

et2

y =

4tet

2

dt= 2et2

+C

y=C et2

2 (3 pts)

y(0) = 1 1 =C 2 C= 3 y(t) = 3et2

2 (1 pt)

3. (10%) dydt

= (y 1)(y 2) , y(0) = 0

1


2/5

Sol:

dy

dt = (y 1)(y 2)

( 1(y 1)(y 2)

)dy = dt (3 pts)

(

1

y 2

1

y 1)dy=

dt (2 pts)

log(|y 2|) log(|y 1|) =t+c

log(

y 2y 1) =t+c

y 2

y 1

=et+c

y =2 c

1et

1 c1et (3 pts for any one of above forms)

y =2 2et

1 2et (Since y(0) = 0) (2 pts)

4. (10%) V V = k , k 2 , , k

+ P(V =k) ( p + , q= 1 p )

Sol:

P(V =k) =Ck11 pq

k2

p= Ck11 p

2

qk2

(10 pts)

Any other answers similar to the correct one receives at most 4 points.

You will get 4 points if your answer has one place different to the correct one.

You will get 2 points if your answer has two places different to the correct one.

You will get 0 point if your answer has more than two places different to the correct one.

5. (20%) X , Y 1 2

P(X= 1, Y= 1) = 4

19 , P(X= 1, Y= 2) =

2

19, P(X= 2, Y= 1) =

7

19 , P(X= 2, Y= 2) =

6

19

P(X= 1) , P(X= 2) , E(X) Var(X) (5%)

Sol:

2


3/5

(a)

P(X= 1) = P(X= 1, Y= 1) + P(X= 1, Y = 2) (1 pt)

= 4

19+

2

19 (2 pts)

= 6

19 (2 pts)

(b)

P(X= 2) = P(X= 2, Y= 1) + P(X= 2, Y = 2) (1 pt)

= 7

19+

6

19 (2 pts)

=13

19

(2 pts)

(c)

E(X) = 1 P(X= 1) + 2 P(X= 2) (1 pt)

= 6

19+ 2

13

19 (2 pts)

=32

19 (2 pts)

(d) Var(X) = E(X2) E(X)2 (1 pt)

E(X2) = 12 6

19+ 22

13

19=

58

19 (2 pts)

Var(X) =58

19 (

32

19)2 =

78

361 (2 pts)

6. (15%) X

f(t) =

ct(1 t) 0 t 1

0

c c , E(X) , Var(X) (5%)

Sol:

Since f(x) is a random function,

1 =

f(x) dx=

10

cx(1 x) dx (2 pts)

=c

10

x x2 dx= c

x2

2

x3

3

10

=c

1

2

1

3

=

c

6

3


4/5

Hence c= 6 (3 pts)

E(x) =

xf(x) dx=

10

x(6x(1 x)) dx (2 pts)

= 6 10 x

2

x3

dx= 6x3

3

x4

41

0

= 6

1

3

1

4

=

1

2 (3 pts)

V(x)

=

(x E(x))2f(x) dx

=E(x2) (E(x))2 (2 pts)

=

x2f(x) dx

1

2

2= 6

10

x3 x4 dx

1

2

2

= 6x4

4

x5

5

1

0

12

2

= 614

1

5 1

2

2

= 1

20 (3 pts)

7. (15%) X , Y

fX(t) =et =fY(t) t 0 , fX(t) = 0 =fY(t) t


5/5

(c) Since X, Yare i.i.d, Cov(X, Y) = 0 and

Var(Z) = Var(X) + Var(Y) + 2Cov(X, Y)

= Var(X) + Var(Y) = 2Var(X)

= 2[E(X2) (E(X))2] (2 pts)

E(X2) =

0

t2et dt= limb

b0

t2et dt

= limb

ett2b0

+ 2 limb

b0

tet = 2

2

So Var(X) = 1

2, and Var(Z) =

2

2 (3 pts)

8. (10%) 30 6 (

Poisson )

Sol:

Let Xbe the random variable of the times of phone calls during 6 seconds. Observe that the

Poisson density is

fX(

x) =

xe

x! x= 0, 1, 2,...

0 otherwise (3 pts)

and = 6 30

60= 3 (3 pts). Hence, we have

P(X 2) = 1 fX(0) fX(1) = 1 e3 3e3 = 1 4e3 (4 pts)

5

Documents

982Bfinalsol