Question

An exponential distribution with unknown parameter λ=θλ=θ is sampled four times, yielding the values 4.1,0.8,0.6,34.1,0.8,0.6,3.

Find each of the following. (Write theta for θ.)

(a) The likelihood function L(θ)=

(b) The derivative of the log-likelihood function =d/dθ[ln⁡L(θ)]=

(c) The maximum likelihood estimate for θ is θ̂ =

Answer 1

If X is a random variable such that $X\sim exp(\theta)$ then the pdf of X is given by--

f(x; в) = ве-т хє *, ө> 0

Then, for a random sample $(x_1,x_2,...,x_n)$ the joint pdf is given by--

$L(\theta)=f(x_1,x_2,..,x_n;\theta)=\prod_{i=1}^{n}[\theta*e^{-\theta x_i}]$

$=\theta^n e^{-\theta \sum_{i=1}^{n}x_i}$(a)

Then, for the random sample of size 7 , the likelihood function is given by--

$L(\theta)=f(x_1,x_2,..,x_7;\theta)=\theta^7 e^{-\theta \sum_{i=1}^{7}x_i}$(a)

log-likelihood function for a random sample $(x_1,x_2,...,x_n)$ is given as--

$lnL(\theta)=ln[f(x_1,x_2,..,x_7;\theta)]=ln[\theta^ n e^{-\theta \sum_{i=1}^{n}x_i}]$

$=n*ln(\theta) - [\theta\sum_{i=1}^{n}x_i ]*ln(e)$

$=n*ln(\theta) - \theta\sum_{i=1}^{n}x_i$

Then, $\frac{\mathrm{d} }{\mathrm{d} \theta}[lnL(\theta)]=\frac{\mathrm{d} }{\mathrm{d} \theta}[n*ln(\theta) - \theta\sum_{i=1}^{n}x_i]$

$=\frac{n}{\theta} -\sum_{i=1}^{n}x_i$(b)

Then, taking $\frac{\mathrm{d} }{\mathrm{d} \theta}[lnL(\theta)]=0$ , we get--

$\frac{n}{\theta} -\sum_{i=1}^{n}x_i=0$

$\Rightarrow \frac{n}{\theta} =\sum_{i=1}^{n}x_i$

$\Rightarrow \hat{\theta}=\frac{n}{\sum_{i=1}^{n}x_i}=\frac{1}{\bar x}$

So, the maximum likelihood function of $\theta$ will be--

$\hat{\theta}=\frac{7}{4.1+0.8+0.6+34.1+0.8+0.6+3}=\frac{7}{6.28571}=0.159$(c)