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If I have an observation $x$ with a Gaussian distributed observational error of standard deviation $\sigma$ then the sum of likelihoods of that observation having the error free values $x_1^{\prime} \dots x_n^{\prime}$ is

$$p(x\mid x^{\prime}_1\dots x^{\prime}_N, \sigma) = \sum^N_{n=1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp \bigg\{-\frac{(x-x_n^{\prime})^2}{2\sigma^2}\bigg\}$$

A simplified but similar example of my problem is as follows. I have a sample of measured heights of men and women, say $(m_1^{\prime} \dots m_n^{\prime})$ and $(w_1^{\prime} \dots w_n^{\prime})$. I then measure the height $x$ of some individual of unknown gender with error $\sigma$ much larger than the errors on my known samples. Replacing $x^{\prime}$ in the above with $m^{\prime}$ and then $w^{\prime}$ I can compare the two sums of likelihoods and determine whether $x$ is more likely to be a male height or a female height.

But what if instead my samples of men and women heights have comparable errors to $x$. Then my question is how to incorporate this additional error.

Thanks

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    $\begingroup$ Your initial statement as it now stands is wrong. "Likelihood" is a standard precisely defined term that means the value of a density or mass function as a function of unobservable parameters with the data fixed. That's not what you have here. You seem to intend this to be a probability density function. That differs from a probability: the value of a density function can be far more than $1$; the value of a probability cannot. Finally, you have a sum where you need a product. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 10 '15 at 17:24
  • $\begingroup$ What I intended to describe above was a sum of likelihoods of an observed value of $x$ having the original value $x^{\prime}_n$ perturbed by a Gaussian distributed observational error. My question is how to incorporate a Gaussian distributed error on $x^{\prime}_n$ i.e. a sum of likelihoods of pairs observations $x$ and $x^{\prime}_n$ being drawn from the same parent population. $\endgroup$ – fen Aug 10 '15 at 20:06
  • $\begingroup$ Your question is opaque. Why would a likelihood be a sum of likelihoods? Do you intend a discrete uniform distribution on the set $\{x_1',\ldots,x_n'\}$? ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 10 '15 at 20:30
  • $\begingroup$ Hi, I have modified the question to (hopefully) better describe my problem. $\endgroup$ – fen Aug 10 '15 at 22:38
  • $\begingroup$ It hasn't become any clearer why you're considering the sum of likelihoods instead of the product. $\endgroup$ – joriki Aug 11 '15 at 15:41

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