MA2216 AY10-11 Sem1

MA2216/ST2131 Probability AY 2010/2011 Sem 1

NATIONAL UNIVERSITY OF SINGAPOREMATHEMATICS SOCIETY

PAST YEAR PAPER SOLUTIONSwith credits to Tan Ah Kow, Koh Ah Tan

MA2216/ST2131 ProbabilityAY 2010/2011 Sem 1

Question 1

(a) Since f(x, y) is a joint p.d.f., ∫∫R2

f(x, y) dxdy = 1.

i.e., ∫ ∞−∞

∫ ∞−∞

Ky−1e−ye− (x−y)2

y dxdy = 1

The LHS is equal to

K

∫ ∞0

y−1e−y∫ ∞−∞

e− (x−y)2

y dxdy

= K

∫ ∞0

y−1e−y√y

2

√2π

∫ ∞−∞

1√y2

√2π

e− 1

2(x−y)2

(√

y/2)2

dxdy

= K√π

∫ ∞0

y12−1e−1y(1) dy

= K√π

Γ(

12

)(1)

12

= K√π√π = Kπ which must be equal to the RHS, which is 1.

Hence, K = 1π . �

(b)

fY (y) =

∫ ∞−∞

f(x, y) dx

=

∫ ∞−∞

Ky−1e−ye− (x−y)2

y dx

= Ky−1e−y∫ ∞−∞

e− (x−y)2

y dx

= Ky−1e−y√y

2

√2π from part (i)

=1√πy−

12 e−y for y > 0. �

Notice that fY (y) = (1)(1/2)

Γ( 12

)y(1/2)−1e−(1)y, y > 0. Thus, Y ∼ Γ(1

2 , 1), from which we also have

IE(Y ) =1/2

1=

1

2and Var(Y ) =

1/2

12=

1

2.

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(c)

fX|Y (x|y) =f(x, y)

fY (y)

=1πy−1e−ye

− (x−y)2

y

1√πye−y

=1√πye− (x−y)2

y for −∞ < x <∞, given Y = y > 0.

Notice that

fX|Y (x|y) =1√

y/2√

2πe− (x−y)2

2(√

y/2)2 .

Hence, the conditional distribution of X, given Y = y > 0, is normal with mean y and variancey2 . �

(d) It follows from the conditional distribution in part (iii) that IE(X|Y = y) = y.Then, using IE(X) = IE[IE(X|Y )], we have IE(X) = IE(Y ) = 1

2 . �

(e) Using the formula Var(T ) = IE(T 2)− [IE(T )]2, we have

IE(X2|Y = y

)= Var(X|Y ) + [IE(X|Y )]2 =

y

2+ y2

With this, we find

IE(X2)

= IE[IE(X2|Y

)]= IE

(Y

2+ Y 2

)=

1

2IE(Y ) + IE

(Y 2)

by linearity

=1

2

(1

2

)+ Var(Y ) + [IE(Y )]2

=1

4+

1

2+

(1

2

)2

= 1

Finally,

Var(X) = IE(X2)− [IE(X)]2 = 1−

(1

2

)2

=3

4. �

Question 2

(a) (Step 1.) First, the transformation is given by{x = 1

2(u− v) (1)

y = 12(u+ v) (2)

(Step 2.) The inverse transformation can be obtained by separately taking (1) + (2) and (1)− (2).These give



{u = y + x (3)

v = y − x (4)

(Step 3.) We find the domain of x and y. The domain of u and v is {(u, v) : u > 0, v > 0}. These,with (1) and (2) respectively, give {

−∞ < x <∞ and y > 0

y > −x and y > x

(Step 4.) The Jacobian

J(u, v) =

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣ 1/2 −1/21/2 1/2

∣∣∣∣ =1

2

The necessary condition of the Jacobian being nonzero for all points in the domain is satisfied.

(Step 5.) Finally, the joint p.d.f. of X and Y is

f(X,Y )(x, y) =1

|J(u, v)|f(U,V )(u, v)

and, noting that Γ(2) = (2− 1)! = 1,

f(U,V )(u, v) = fU (u)fV (v) (since U and V are independent)

=(1/2)2

Γ(2)u2−1e−

12u · (1/2)2

Γ(2)v2−1e−

12v

=1

16uve−

12

(u+v)

Therefore,

f(X,Y )(x, y) =1

|J(u, v)|f(U,V )(u, v)

=1

|1/2|1

16uve−

12

(u+v)

=1

8(y + x)(y − x)e−y from (2), (3) and (4)

=1

8

(y2 − x2

)e−y for −∞ < x <∞, y > |x|. �

(b)

Cov(X,Y ) = Cov

(1

2U − 1

2V ,

1

2U +

1

2V

)=

1

2· 1

2Cov(U − V , U + V ) (by bilinearity)

=1

4[Cov(U,U) + Cov(U, V )− Cov(V,U)− Cov(V, V )] (by bilinearity)

=1

4[Var(U) + 0− 0−Var(V )] (since U and V are independent)

= 0

(Var(U) = Var(V ) since U and V are identically distributed.) �



(c) No. The domain of x and y is {(x, y) : −∞ < x <∞, y > |x|}. This shows that y depends on x.�

Question 3

(a) We know that for an indicator random variable IA of an event A, IE(IA) = IP(A).Hence, for each i,

IE(Xi) = IP{ball ri is withdrawn} =

1 ·(

2911

)(

3012

) = 0.4 �

Similarly, for each j,

IE(Yj) = IP{jth ball is withdrawn} =

1 ·(

2911

)(

3012

) = 0.4 �

Since Xi and Yj are independent,

IE(XiYj) = IE(Xi) IE(Yj) = (0.4)(0.4) = 0.16. �

Cov(X,Y ) = Cov

10∑i=1

Xi ,8∑j=1

Yj

=

10∑i=1

8∑j=1

Cov(Xi, Yj) (by bilinearity)

=

10∑i=1

8∑j=1

0 (since Xi and Yj are independent)

= 0. �

(b) Notice that X and Y are indicator random variables.Let A and B be, respectively, the events that X and Y represent, i.e.,

X =

{1, if A occurs

0, if A does not occurY =

{1, if B occurs

0, if B does not occur

To show that “X and Y are independent ⇐⇒ Cov(X,Y ) = 0” :

The direction “=⇒” is true for any random variables X and Y . For the direction “⇐=”,

Cov(X,Y ) = 0

⇒ IE(XY )− IE(X) IE(Y ) = 0

⇒ IP(AB)− IP(A) IP(B) = 0

⇒ IP(AB) = IP(A) IP(B)

i.e., A and B are independent. Hence, X and Y are independent. �



Question 4

(a) We have independent X1, X2, · · · ∼ U(0,1) and N ∼ Geom(p). Let q = 1− p.Also, notice that M takes values between 0 and 1. Thus, for 0 ≤ x < 1,

IP(M ≤ x) =∞∑n=1

IP{M ≤ x|N = n} · IP{N = n}

=

∞∑n=1

IP{max(X1, X2, · · · , Xn) ≤ x} · qn−1p

=∞∑n=1

IP{X1 ≤ x,X2 ≤ x, · · · , Xn ≤ x} · qn−1p

=∞∑n=1

IP{X1 ≤ x} IP{X2 ≤ x} · · · IP{Xn ≤ x} · qn−1p (since X1, X2, · · · are independent)

=

∞∑n=1

IP{X1 ≤ x}n · qn−1p (since X1, X2, · · · are identically distributed)

=∞∑n=1

(x− 0

1− 0

)n· qn−1p for 0 ≤ x < 1

= px

∞∑n=1

(qx)n−1

= px · 1

1− qx(|qx| < 1 since q, x ∈ (0, 1))

=px

1− (1− p)xfor 0 ≤ x < 1

We can thus deduce that

IP(M ≤ x) =

0, for x < 0

px1−(1−p)x , for 0 ≤ x < 1

1, for x ≥ 1

�

(b) We first calculate

IE(Sn) = IE(X1 +X2 + · · ·+Xn)

= IE(X1) + IE(X2) + · · ·+ IE(Xn) (by linearity)

= n IE(X1) (since X1, X2, · · · , Xn are identically distributed)

= n IP(X1 = 1) (since X1 is an indicator random variable)

= np

and

Var(Sn) = Var(X1 +X2 + · · ·+Xn)

= Var(X1) + Var(X2) + · · ·+ Var(Xn)

(since X1, X2, · · · , Xn are independent, the covariances vanish)

= nVar(X1) (since X1, X2, · · · , Xn are identically distributed)

= np(1− p)



Thus, for large n, by CLT, Sn ∼ N(np, np(1− p)) approx. Equivalently,

Sn =Snn∼ N

(p,p(1− p)

n

)approx.

(i.)

IP{∣∣Sn − p∣∣ ≥ c} = IP

∣∣∣∣∣∣ Sn − p√

p(1−p)n

∣∣∣∣∣∣ ≥ c√p(1−p)n

≈ IP

|Z| ≥ c1/2√900

(∵√p(1− p) ≈ 1/2

)= IP {|Z| ≥ 60c}= 2 IP{Z ≥ 60c} (by symmetry)

Thus,

IP{Z ≥ 60c} ≈ 1

2IP{∣∣Sn − p∣∣ ≥ c} =

1

2(0.01) = 0.005

60c ≈ z0.005 = 2.58

c ≈ 0.043 �

(ii)

IP{∣∣Sn − p∣∣ ≥ 0.025

}= IP

∣∣∣∣∣∣ Sn − p√

p(1−p)n

∣∣∣∣∣∣ ≥ 0.025√p(1−p)n

≈ IP

|Z| ≥ 0.0251/2√n

= IP

{|Z| ≥ 0.05

√n}

= 2 IP{Z ≥ 0.05

√n}

(by symmetry)

Thus,

2 IP{Z ≥ 0.05√n} ≈ IP

{∣∣Sn − p∣∣ ≥ 0.025}

= 0.01

IP{Z ≥ 0.05√n} ≈ 0.005

0.05√n ≈ z0.005 = 2.58

n ≈ (51.6)2 ≈ 2663 �


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MA2216 AY10-11 Sem1