# Integrate Gaussian PDF over another gaussian, to get probability of imprecise datapoint

I have a Gaussian PDF and an observation datapoint. To get a probability, I cannot use the datapoint itself--I need a range around it. But say the datapoint is imprecise and obeys a normal distribution, whose mean and variance I know. How can I integrate the PDF over the datapoint's mean and variance?

I think I get the gist of the necessary integral. For each piece of the domain, integrate the main PDF (get the likelihood it falls in that little piece), integrate the datapoint PDF (get the likelihood the datapoint falls in that piece), multiply them, and integrate over all the pieces. What is the solution to this integral?

EDIT: Let me give the full context. I am investigating automatic speech recognition. It uses Gaussian Mixture Models to calculate the probability that a feature vector (39 numbers that characterize a 10ms audio frame) would be produced by a phoneme. Essentially, the feature vector is input to a 39-dimensional multivariate Gaussian. However, the value is used directly. No integration is done. Therefore it's not a probability at all, which is a problem because other algorithms (eg Hidden Markov Models) expect probabilities. (Quite suprising how this cludge is central to ASR.) I am trying to figure out what I can do to get real probabilities. Treating the feature vector as also a multivariate gaussian (or a n-dimensional cube with sides equal to the variances) is a way of giving it some volume to integrate over.

• You may want to be more explicit in describing the problem, but I think you are describing a convolution integral of two normal distributions, which ends up being a normal distribution in which the means and variance is the sum of the means and variances, respectively, of the original normal distrbutions. – Ron Gordon Jan 30 '13 at 13:12
• I am not sure, whether I understood your question correctly. If you mean that you have two random variables $X$ and $Y$ where $Y$ represents somehow a datapoint, and $X$ is normally distributed conditional on $Y$ - you just use the law of the total probability. – Ilya Jan 30 '13 at 13:23