# Likelihood for a set of $N$ data points

Refering from , given a set of $$N$$ data points $$D = \{x_i\}$$ drawn independently from a Gaussian with mean $$\mu$$ and standard deviation $$\sigma$$, the likelihood for these data is $$P(D|\mu, \sigma) = \prod_{i = 1}^{N}\dfrac{1}{\sigma \sqrt{2\pi}}\exp{\bigg[ -\dfrac{(x_i - \mu)^2}{2\sigma^2}\bigg]}.$$

How is this equation equivalent to

$$P(D|\mu, \sigma) = \dfrac{1}{\sigma^N {(2\pi)}^{N/2}}\exp{\bigg[ -\dfrac{1}{2\sigma ^2} \sum_{i=1}^{N}(x_i - \mu)^2\bigg]}?$$

Is there a simple proof that these two expressions are equivalent?

REFERENCE

 Bailer-Jones, C. (2017), Practical Bayesian Inference: A Primer for Physical Scientists, Cambridge: Cambridge University Press, p. 126.

Is there a simple proof that these two expressions are equivalent?

Simply using basic algebraic properties of the powers

Example

$$\prod_{i=1}^N \sigma=\underbrace{\sigma\cdot\sigma\dots\sigma}_{\text{N times}}=\sigma^N$$

$$\prod_{i=1}^N e^{x_i}=e^{x_1}\cdot e^{x_2}\dots e^{x_N}= e^{\sum_i x_i}$$

Is this enough to clarify your dubts?

• Thanks, yes indeed my doubts are clarified. Jun 15, 2021 at 12:11