Question

Question 4 16 marks Let Y N(Hy, o). Then X := exp(Y) is said to be lognormally distributed with p.d.f. (In(x)-Hy) exp 202 fx(x) TOYV27 and denoted as LN(Hy, of). Let Xı,... , X, be random samples from the LN(Hy,of) distribution (a) Find the maximum likelihood estimator for ty, which we denote as fty (Hint: Use the fact that Yi In(X) is normally distributed with known mean and variance) Verify that the sought stationary point is a maximum (b) Verify that the estimator y is unbiased and derive its variance. Is the estimator consistent? (c) Assuming y is normally distributed, how many data points would you need to collect to obtain a 95% random interval of y 1 for py if of = 2? (d) It is straightforward to show that xE(X) exp(y +o%/2). Verify that although y is an unbiased estimator of y, exp(ty +o3/2) is not an unbiased estimator of x, for n 1

Answer 1

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as for given data...

Let Y ~N(rly, σ). Then X := exp(Y) is said to be lognormally distributed with pdf. (In(x)y)2 202 and denoted as LN(μΥ, σ) Let Xi, ,X, be random samples from the LN(py, σ distribution

EXPLANATION ::-

We have

(In (x)-py)2 Σ; 0<x< o),i = 1,2, 1 fx(x) η e χ/2πσγ

(a) Find the maximum likelihood estimator for μΥ, which we denote as μΥ (Hint: Use the fact that Y In(X) is normally distributed with known mean and variance) Verify that the sought stationary point is a maximum.

SOL ::-

a)The joint PDF is

fxX(X, X2,Xn) II fx(x) i-1 (In(X)-y)2 1 fx1,X2...Xn(X1, X2, Xn)T, X 2Toy 1 e

The loglikelihood function is

ΣL)-μγ 2 Σ Π 1 1 L (X1, X2, , Χ) , μY, σγ) ε Ε ΠΕΧν2πσγ (Χ1, X , . ., Χ), μγ, σy) - InL (X, X , Χ i μγ, σγ) - In | Χ-n In (2πσ,)- ΣΗln(X)-Y)2 2σ Π 1 (Χ1, X2 , X ; μY, σγ) - i-1

Taking partial derivatives and equating to 0,

- (Χ, X2, . , Χ) μΥ, σy) - Αίρ ,Σ.in(X)-μ)-0 20φ Σ- In(X) 24 μΥ

The second derivative
-n I(X1, X2,., Xni y, ay) 2 202

Hence the sought point is maximum.

(b) Verify that the estimator consistent? is unbiased and derive its variance. Is the estimator

SOL ::-

1 In(X) E (y) E i-1 n Y E (fty) = E ( E (Y)HY i-1 n

Hence
μΥ'
is unbiased.

In(X) Var (Ry) Var ( ΣΥ Var (y) Var (Li=1. Var (y)

(c) Assuming fy is normally distributed, how many data points would you need to collect to obtain a 95% random interval ofPv t l for μΥ if σ '-2?

SOL ::-

We need the margin of error

OY 1 ME 1.96 2 Y n 1.962 n> 1.962(4) n> 16

(d) It is straightforward to show that Ax := E(X) = exp(pv + σ/2). Verify that although μΥ is an unbiased estimator of μΥ, exp(pv + σ/2) is not an unbiased estimator of Ax, for n 1

SOL ::-

For $n=1$ ,
LyIn(X) 1.
.

$\widehat{\mu_Y }=\ln(X_1)\\ E(e^{\widehat{\mu_Y }+\sigma ^2_Y/2})=E(e^{\ln(X_1)+\sigma ^2_Y/2})\\ E(e^{\widehat{\mu_Y }+\sigma ^2_Y/2})=E\left (X_1 \right )E(e^{\sigma ^2_Y/2})\\ E(e^{\widehat{\mu_Y }+\sigma ^2_Y/2})=\mu _Xe^{\sigma ^2_Y/2}\neq \mu _X$

Hence
eytof/2
is not an unbiased estimator of $\mu _X$

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