Question

A random sampleY1 , Y 2 , ⋯,Yn is drawn from a distribution whose probability density function is given by: f (Y ) = βe − βY , Y  0 & β > 0 a). Obtain the maximum likelihood estimator (MLE) of β. (3 points) 1Econometrics- Assignment I b). Given that ∑ n Y i = 25 , ∑n Yi 2 = 50 , n = 50 calculate the maximum likelihood i =1 i =1 estimate of β. (3 point) c). Using the same data as in part (b), test the null hypothesis that β =1against the alternative hypothesis that β ≠1at 5% level of significance

Answer 1

(a)From the given information, the probability density function is,

$f(y)=βe^{−βy}, y>0, β>0 .$

Consider, the likelihood function,

$L(y)=∏^{n}_{i=1}f(y) \\ =f(y1)*f(y2)*.....*f(y_n) \\ =βe^{−βy1}*βe^{−βy2}*............*βe^{−βy_n } \\\ =β^{n}e^{−β}∑yi$

Consider, ln on both sides

$lnL(y)=ln[β^{n}e^{−β}∑y_i] \\ =nlnβ−β∑yi$

Therefore,

$\frac{dlnL(y)}{dβ}=0\\\frac{n}{β}−∑Yi=0 \\ \frac{ n}{β}=∑Yi \\ βˆ=\frac{1}{y}$

$b)\\ ∑ Y i = 25 , ∑ Yi^2= 50 , n = 50 \\ Therefore,\\ βˆ=\frac{1}{y}\\=\frac{1}{\frac{25}{50}} \\ =2$

(c)null hypothesis that β =1against the alternative hypothesis that β ≠1at 5% level of significance.

Null Hypothesis:

H0: β =1

Alternative Hypothesis:

H1: β ≠1

Consider, the likelihood ratio,

$\frac{L(y )H_0}{L(y)H_1}=\frac{1^{n}e−1∑y_i}{β^{n}e^{−β}∑y_i }\\ =(\frac{1}{β})^{n}e^{(β−1)∑y}\\ln\frac{L(y )H_0}{L(y)H_1}=nln(\frac{1}{β})+(β−1)∑y$

Critical region is,

$ln\frac{L(y )_{H_0}}{L(y)_{H_1}}≤c\\nln(\frac{1}{β})+(β−1)∑y≤c\\(β−1)∑y≤c−nln(\frac{1}{β}) \\ ∑y≤\frac{c−nln(\frac{1}{β})}{(β−1) }\\ ∑y≤c',\frac{c−nln(\frac{1}{β})}{(β−1)}=c'$

Thus, reject the null hypothesis if, $∑y≤c'$