Question

A random sample of size $n$ from the normal distribution has the density $\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma}} e^{-(1/2\sigma^2)(x_i-\mu)^2} = \left(\frac{1}{2\pi\sigma^2}\right)^{n/2} \exp \left[ -\frac{1}{2\sigma^2} \sum (x_i - \mu)^2 \right]$. Find Maximum Likelihood.

          A random sample of size $n$ from the normal distribution has the density
$\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma}} e^{-(1/2\sigma^2)(x_i-\mu)^2} = \left(\frac{1}{2\pi\sigma^2}\right)^{n/2} \exp \left[ -\frac{1}{2\sigma^2} \sum (x_i - \mu)^2 \right]$.
Find Maximum Likelihood.

A random sample of size n from the normal distribution has the density
∏i=1^n(1)/(√(2πσ)) e^-(1/2σ^2)(xi-μ)^2 = ((1)/(2πσ^2))^n/2[ -(1)/(2σ^2)∑ (xi - μ)^2 ].
Find Maximum Likelihood.

Added by Luke W.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Adriano Chikande

04/11/2022

Step 1

Step 1: The likelihood function is the product of the probability density functions of each observation. Show more…

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A random sample of size n from the normal distribution has the density $$ \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma}} e^{-(1/2\sigma^2)(x_i-\mu)^2} = \left(\frac{1}{2\pi\sigma^2}\right)^{n/2} exp\left[-\frac{1}{2\sigma^2}\sum(x_i-\mu)^2\right]. $$ Find Maximum Likelihood.