Question

Let the probability density function of the continuous random variable 𝑋 be
, 0, 𝑥 ≥ 0 otherwise , 
where 𝜆 > 0 is a parameter. If the observed sample values of 𝑋 are 
𝑥₁ = 1.75, 𝑥₂ = 2.25, 𝑥₃ = 2.50, 𝑥4 = 2.75, 𝑥₅ = 3.25, then the Maximum Likelihood Estimator of 𝜆 is

• A. \(\frac{5}{2}\)
• B. \(\frac{1}{5}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{2}{5}\)

Answer 1

The Correct Option is D

To find the Maximum Likelihood Estimator (MLE) of \(\lambda\) for the given probability density function (PDF) and observed sample values, we need to consider the exponential distribution. The probability density function is given as:

The PDF of an exponential distribution is defined as: 

\(f_X(x, \lambda) = \lambda e^{-\lambda x}, \quad x \geq 0\text{ and } \lambda > 0\)

For independent and identically distributed samples \(x_1, x_2, \ldots, x_n\), the likelihood function \(L(\lambda)\) is:

\(L(\lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i}\)

Taking the natural logarithm of the likelihood function, we get the log-likelihood function:

\(\ln L(\lambda) = n \ln \lambda - \lambda \sum_{i=1}^{n} x_i\)

To find the MLE of \(\lambda\), differentiate the log-likelihood function with respect to \(\lambda\) and set it to zero:

\(\frac{d}{d\lambda}(\ln L(\lambda)) = \frac{n}{\lambda} - \sum_{i=1}^{n} x_i = 0\)

Solving for \(\lambda\), we get:

\(\lambda = \frac{n}{\sum_{i=1}^{n} x_i}\)

Given the sample values: \(x_1 = 1.75, x_2 = 2.25, x_3 = 2.50, x_4 = 2.75, x_5 = 3.25\), we first calculate \(\sum_{i=1}^{5} x_i\):

\(\sum_{i=1}^{5} x_i = 1.75 + 2.25 + 2.50 + 2.75 + 3.25 = 12.50\)

Now substitute into the formula for \(\lambda\):

\(\lambda = \frac{5}{12.5} = \frac{2}{5}\)

Therefore, the Maximum Likelihood Estimator of \(\lambda\) is \(\frac{2}{5}\).