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Nina [5.8K]
3 years ago
10

Let X1, X2, ... , Xn be a random sample from N(μ, σ2), where the mean θ = μ is such that −[infinity] < θ < [infinity] and

σ2 is a known positive number. Show that the maximum likelihood estimator for θ is θ^ = X.
Mathematics
1 answer:
Sliva [168]3 years ago
6 0

Answer:

l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)

And then the maximum occurs when l'(\theta) = 0, and that is only satisfied if and only if:

\hat \theta = \bar X

Step-by-step explanation:

For this case we have a random sample X_1 ,X_2,...,X_n where X_i \sim N(\mu=\theta, \sigma) where \sigma is fixed. And we want to show that the maximum likehood estimator for \theta = \bar X.

The first step is obtain the probability distribution function for the random variable X. For this case each X_i , i=1,...n have the following density function:

f(x_i | \theta,\sigma^2) = \frac{1}{\sqrt{2\pi}\sigma} exp^{-\frac{(x-\theta)^2}{2\sigma^2}} , -\infty \leq x \leq \infty

The likehood function is given by:

L(\theta) = \prod_{i=1}^n f(x_i)

Assuming independence between the random sample, and replacing the density function we have this:

L(\theta) = (\frac{1}{\sqrt{2\pi \sigma^2}})^n exp (-\frac{1}{2\sigma^2} \sum_{i=1}^n (X_i-\theta)^2)

Taking the natural log on btoh sides we got:

l(\theta) = -\frac{n}{2} ln(\sqrt{2\pi\sigma^2}) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i -\theta)^2

Now if we take the derivate respect \theta we will see this:

l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)

And then the maximum occurs when l'(\theta) = 0, and that is only satisfied if and only if:

\hat \theta = \bar X

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