Suppose I have two normal distributions of measurement taken with different instruments. What is the best strategy to calculate weighted mean of two distributions if error distributions of both the instruments is unknown? Is it possible?


2 Answers 2


The sum of two independent normal variables is normal random variable, e.g. $x\sim\mathcal{N}(\mu_x,\sigma_x^2)$ and $y\sim\mathcal{N}(\mu_y,\sigma_y^2)$ will get you $$\alpha x+(1-\alpha)y\sim\mathcal{N}(\alpha\mu_x+(1-\alpha)\mu_y,\alpha^2\sigma_x^2+(1-\alpha)^2\sigma_y^2)$$ Here, you could use $\alpha=\frac{1}{2}$ for an equal weight mean.

If you assume that both instruments are unbiased, then you actually have a simpler situation: $$x\sim\mathcal{N}(\mu,\sigma_x^2)$$ $$y\sim\mathcal{N}(\mu,\sigma_y^2)$$

In this case you assume that in average both instruments are accurate(as defined by IUPAC), i.e. have no bias. However, their precision is different $\sigma_x,\sigma_y$.

Let's construct a weighted estimator $$\hat\mu=\alpha x + (1-\alpha) y$$

Let's look at its characteristics: $$E[\hat\mu]=\alpha\mu+(1-\alpha)\mu=\mu $$

Good, it's unbiased regardless of the weight $\alpha$, i.e. it's accurate.

Let's see what's its precision: $$Var[\hat\mu]=\alpha^2\sigma_x^2+(1-\alpha)^2\sigma_y^2$$

The independence assumption of normal variables is usually reasonable for instrument measurements unless they're affected with the same exact random shocks, which may happen in certain setups but not usually encountered.

In this case the the optimal $$\alpha=\frac{\sigma_y^2}{\sigma_x^2+\sigma_y^2}$$

You can see that if the precisions are the same, the weight is $\alpha=1/2$. Otherwise, if the first instrument twice more precise, e.g. $\sigma_x=\sigma_y/2$ then you get $$\alpha=\frac{4}{4+1}=0.8$$

  • $\begingroup$ Isn't there any way to assign weights to μx and μy based on variances of those distributions? Because in above case we are assuming that both of the measurement instrument are equally accurate $\endgroup$
    – Siddhesh
    Oct 29, 2015 at 13:49
  • 1
    $\begingroup$ @Siddhesh, no, we're not assuming that they're equally accurate as you can see we have $\sigma_x,\sigma_y$. You can assign weights but what is your objective? $\endgroup$
    – Aksakal
    Oct 29, 2015 at 13:52
  • $\begingroup$ @Aksakal you could also add that if X is normal, than aX+b (where a and b are const.) is also normal what makes the math easy and applicable. $\endgroup$
    – Tim
    Oct 29, 2015 at 14:02
  • $\begingroup$ The sum of two normals is normal if and only if they are marginals of bivariate normal distribution. Independence usually ensures that, but if the variables are not independent their sum might not be normal. $\endgroup$
    – mpiktas
    Oct 29, 2015 at 14:34
  • $\begingroup$ @mpiktas, thanks for correction, I updated the answer $\endgroup$
    – Aksakal
    Oct 29, 2015 at 14:52

I will edit this answer into a more elaborate one later in the day.

You can consider the geodesic between your two densities and pick-up the distribution at the mid-distance. These densities have an hyperbolic geometry under the Fisher-Rao metric. You can google SIR Costa Information Geometry for detailed computations and close-form expressions.

  • $\begingroup$ What properties does this have as an estimator? In other words, what characteristics of this answer actually address the question? $\endgroup$
    – whuber
    Dec 9, 2017 at 16:40

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