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I am trying to find out the the Radon-Nikodym derivative of $μ_1$ w.r.t $μ_2$ where $μ_1$ is a Gaussian measure on $R^n$ with mean $a$, and standard deviation $σ$, and $μ_2$ is also a Gaussian measure, with mean $b$ and standard deviation $σ$.

Solution Attempt

Let $n\in N$ and let $B_0(R^n)$ denote the completion of the Borel $\sigma$-algebra on $R^n$. Let $\lambda^n:B_0(R^n)\rightarrow[0,R^+]$ denote the usual $n$-dimensional Lebesgue measure.

Based on the definition of Gaussian measure and it's properties we know the following two equivalent relations $\lambda^{n} \ll \mu_1^{n} \ll \lambda^{n}$ and $\lambda^{n} \ll \mu_2^{n} \ll \lambda^{n}$ where $\ll$ stands for absolute continuity of measures; Hence we have absolute continuity relations $\mu_1^{n} \ll \lambda^{n}$ and $\lambda^{n} \ll \mu_2^{n}$.

Also since $\lambda^{n} \ll \mu_1^{n} \ll \lambda^{n}$ we can use the Radon-Nikodym property $\dfrac {d\lambda } {d\mu_1}=\left( \dfrac {d{\mu_1}} {d\lambda }\right) ^{-1}$

Since both Lebesgue and Gaussian measures are $\sigma$-finite and based on the above relations we should be apply the Radon-Nikodym Chain Rule here. $$\dfrac {d\mu_2 } {d\mu_1} = \dfrac {d\mu_2 } {d\lambda}\dfrac {d\lambda } {d\mu_1} = \dfrac {d\mu_2 } {d\lambda}\left( \dfrac {d{\mu_1}} {d\lambda }\right) ^{-1}$$

The Gaussian measures $μ_1:B_0(R^n)\rightarrow[0,R^+]$ and $μ_2:B_0(R^n)\rightarrow[0,R^+]$ are defined as

${\mu_1}(E) = (2\pi{\sigma}^2)^\frac{-n}{2}\int _{E}exp\left( -\dfrac {1} {2\sigma ^{2}}\left\| x-a\right\| _{R_{n}}^{2}\right)d\lambda ^{n}\left( x\right) $

and ${\mu_2}(E) = (2\pi{\sigma}^2)^\frac{-n}{2}\int _{E}exp\left( -\dfrac {1} {2\sigma ^{2}}\left\| x-b\right\| _{R_{n}}^{2}\right)d\lambda ^{n}\left( x\right) $ respectively.

I am unsure though how to proceed further here, any help would be much appreciated.

Thanks in advance.

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up vote 1 down vote accepted

$$\frac{\mathrm d\mu_1}{\mathrm d\mu_2}(x)=\exp\left(-\frac1{2\sigma^2}(\|x-a\|^2-\|x-b\|^2)\right)=\exp\left(\frac{\langle a-b,x-\frac12(a+b)\rangle}{\sigma^2}\right) $$

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Hi Mate Thanks for your answer. I was also hoping, if you would be kind enough to give me a clue how to evaluate the same derivative if we did not know the standard deviations but had a covariance matrix instead. – Comic Book Guy Oct 3 '12 at 23:05

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