Question 1

(a)  [6 Marks] Let’s assume that X1, X2 , ..., Xn   f (x; θ), where the density is given by,

f (x; θ) =

when, 0 < xi  < θ and f (x; θ) = 0 otherwise. Let’s assume we have the following estimator of θ θˆ = 2

Find the Mean Squared Error (MSE) of θˆ.

Solution

Here we can see that E(θˆ) = E(2) =  × ∑ E(Xi).

It can be shown that E(Xi) =  .  Thus, E(θˆ) = n   = θ .  Thus the bias E(θˆ) − θ = 0 The

Var(θˆ) = Var(2)

=  Var(Xi);

4n θ2

=

n2  12

2

=

3n

The MSE is calculated by,

MSE(θˆ) = (E(θˆ) − θ = 0)2 + Var(θˆ) =

(b)  [4 Marks] Using the information in part (a) above, is T = θˆ2  = 42  an unbiased estimator of ψ(θ) = θ 2 ?  If yes, prove your answer.  If not find the bias of T  (i.e.  Find Biasθ(T)) as an estimator of ψ(θ) = θ 2 .

Solution

E(T) = E(θˆ2 ) =  (E(θˆ))2  + V (θˆ) = θ 2  +    θ 2  and so T is not an unbiased estimator of ψ(θ) = θ 2 .

Biasθ(T) = E(T) − ψ(θ) = (θ 2 + ) − θ 2  =       ■

Question 2

Let’s assume that X1, X2 , ..., Xn  ∼ f (x; µ, σ), where the density is given by,

f (x; µ; σ) =  exp (− (ln(x) − µ)2 ) ,  0 < x < ∞

(a)  [4 Marks] Find the maximum likelihood estimator for µ

Solution: The likelihood function,

L(µ | s) =   exp (− (ln(xi) − µ)2 )

ℓ(µ | s) ∝ −  (∑ (ln(xi) − µ)2

ℓ′ (µ) =  ∑ (ln(xi) − µ) = 0

⇒  = ∑ ln(xi)

(b)  [6 Marks] Assume σ = 1. Let τ be the 95th percentile (i.e.,  f (x; µ, σ)dx = 0.95). Find the value of τ as a function of µ . That is find τ = ψ(µ). Using the relationship ψ(.), find the MLE of τ .

Solution: Here, X ∼ lognormal(µ = σ = 1). Thus, ln(X) ∼ N(µ, 1). Then,

P (X < τ ) = 0.95

⇒P (ln(X) < ln(τ )) = 0.95

⇒P (ln(X) − µ < ln(τ ) − µ) = 0.95

⇒P (Z < ln(τ ) − µ) = 0.95

This implies, that ln(τ ) − µ = 1.645 ⇒ τ = exp(µ + 1.645). Using the invariance property of

Question 3

Suppose we have a finite population Π and a measurement X  : Π → {0, 1}, where |Π| = 15 and |{π : X(π) = 0}| = 4.

(a)  [2 Marks] Determine fX (0) and fX (1).

(b)  [4 Marks] For an i.i.d. sample of size 5 (this means sampling with replacement), determine the probability that 5fˆX (1) = 2.

Solution

Note that 5fˆX (1) is the number of 1’s in the sample of size 5 and under iid sampling 5fˆX (1) ∼

Bin(n = 5, θ = 11/15) and so P (5fˆX (1) = 2) = (2(5)) ( )2  (1 − )3        ■

(c)  [4 Marks] For a simple random sample of size 5 (without replacement), determine the probability that 5fˆX (1) = 2.

Solution

Let X be the number the number of elements with measurement 1 in a sample of size 5. Then, P (5fˆX (1)  =  2)  ⇒ P (fˆX (1)  =  2/5).  That is we need to find P (X  =  2) with replacement. Implying hypergeometric rule we get,

( 2(11))(3(4))

( 5(15))

■

Question 4

A random variable X  has the following distribution with parameter θ  >  0, and the PDF is as follows,

f (x; θ) =  exp (− ) ;  x > 0

(a)  [5 Marks] If we are interested in making inference about the median of the distribution then determine ψ(θ)

Hint: The median M is calculated as  f (x; θ)dx = 0 .5. Then solve for M

Solution: Calculating the CDF,

F (x; θ) =   exp (− )

=  z exp(−z)dz;         [substitution]

= 1 − exp( −z) = 1 − exp (− )

Let the median be M . Thus,

1 − exp (− ) = 0.5

⇒ −  = ln(0.5) = − ln(2)

⇒M2  = 2θ2 ln(2)

⇒M = θ √2 ln(2)

The median is θ√2 ln(2).

(b)  [5 Marks] If we are interested in making inference about the mode of the distribution then determine ψ(θ). You don’t need to perform the second derivative test.

Hint:  The mode of X  is calculated as by differentiating the PDF (or ln of the PDF) with respect to X and then find the maxima.

Solution: Take the ln of the PDF,

ln(f (x; θ)) = ln(x) − ln(θ2 ) −

∂ ln(f (x; θ))       1       2x

∂x               x      2θ2

1        x

=

x      θ 2

x2  = θ 2 x = θ;

The mode is θ .

since,x > 0