Question

3. (6 pts) Suppose \( X_{i} \sim N\left(0, a_{i} \theta\right) \) independently for \( i=1,2, \ldots, n \) where \( a_{i}(>0) \) are fixed and known constants for all \( i \). Find the MLE of \( \thet

Answer 1

Final Answer:

The maximum likelihood estimator
`(MLE) of \( \theta \) is \( \hat{\theta} = (n)/(\sum_(i=1)^(n)a_(i)) \).`

Step-by-step explanation:

The given scenario describes independent random variables
`\( X_(i) \)`with a normal distribution
`\( N(0, a_(i)\theta) \)`for
`\( i=1,2,\ldots,n \).` Here,
`\( a_(i) \)` are known constants.

The probability density function (PDF) of a normal distribution is
`\( f(x) = (1)/(√(2\pi)\sigma)e^{-(1)/(2)\left((x-\mu)/(\sigma)\right)^(2)} \)`, where
`\( \mu \)`is the mean and
`\( \sigma \)`is the standard deviation.

In the given case,
`\( X_(i) \)` follows
`\( N(0, a_(i)\theta) \)`, so the PDF is
`\( f(x) = \frac{1}{\sqrt{2\pi a_(i)\theta}}e^{-(1)/(2)\left(\frac{x}{\sqrt{a_(i)\theta}}\right)^(2)} \).`

The likelihood function is the product of the PDFs of all
`\( X_(i) \)'s,` given by
`\(`
`L(\theta) = \prod_(i=1)^(n)f(x_(i)) \).`

To find the MLE, we maximize the likelihood function with respect to
`\( \theta \)`, which is equivalent to maximizing the log-likelihood function. After some algebraic manipulations, the MLE of
`\( \theta \)`is found to be
`\( \hat{\theta} = (n)/(\sum_(i=1)^(n)a_(i)) \).`

This result makes intuitive sense as it suggests that the MLE of
`\( \theta \)`is inversely proportional to the sum of the known constants
`\( a_(i) \)`, reflecting the contribution of each observation to the overall estimate.