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Refering from [1], 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

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

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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?

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

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