Suppose $X$ is normally distributed with mean $10$ and standard deviation $1$. I take a normally-distributed noisy measurement of $X$ with standard deviation $0.1$, and the measurement is $5$. I am interested in determining the posterior distribution of $X$. Is there a general rule for this?

Thinking about it, I see this is equivalent to: $$X \sim N(10,1) \\ Y \sim N(X,0.1)$$ and then asking $P(X | Y = 5)$, which requires solving: $$P(X=x | Y=5) = \frac{f(x,5,0.1) f(x,10,1)}{ \int_{x'} f(x',5,0.1) f(x',10,1) dx'}$$ where $f$ is the normal density, but that's how far I got.

  • $\begingroup$ Standard result. Somewhat messy algebra but conceptually simple application of Bayes' Thm. Result is another normal. Result often simplified by using 'precision' $\tau = 1/\sigma^2$ throughout. Perhaps see these notes. $\endgroup$
    – BruceET
    Dec 25, 2018 at 16:28

1 Answer 1


For more general situation, let $X\sim N(\mu,\sigma_X^2)$ and noise $\epsilon \sim N(0,\sigma_\epsilon^2)$. Then the measurement $Y=X+\epsilon$.

$$\left( \begin{matrix} X\\Y\end{matrix} \right) = \left( \begin{matrix} 1 &0 \\ 1& 1\end{matrix} \right) \left( \begin{matrix} X \\ \epsilon \end{matrix} \right)$$ Then $\left( \begin{matrix} X\\Y\end{matrix} \right)$ follows the bivariate normal distribution with mean and variance matrix $$E\left( \begin{matrix} X\\Y\end{matrix} \right) = \left( \begin{matrix} \mu\\ \mu\end{matrix} \right)$$

$$Var\left( \begin{matrix} X\\Y\end{matrix} \right) = \left( \begin{matrix} 1 &0 \\ 1& 1\end{matrix} \right) \left( \begin{matrix} \sigma_X^2 &0 \\ 0&\sigma_\epsilon^2\end{matrix} \right) \left( \begin{matrix} 1 & 1 \\ 0& 1\end{matrix} \right) = \left( \begin{matrix} \sigma_X^2 & \sigma_X^2 \\ \sigma_X^2& \sigma_X^2+\sigma_\epsilon^2\end{matrix} \right) $$

The distribution of $(X|Y)$ also follows normal distribution with mean and variance: $$E(X|Y) = \mu +\frac {\sigma_X^2}{\sigma_X^2+\sigma_\epsilon^2}(Y-\mu)$$ $$Var(X|Y) = \sigma_X^2 - \frac {\sigma_X^4}{\sigma_X^2+\sigma_\epsilon^2}$$

  • $\begingroup$ Thank you! Could you please indicate the general result used to derive E(X|Y) and Var(X|Y)? It would make the answer even more useful. $\endgroup$
    – user118967
    Dec 26, 2018 at 20:20

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