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We have several liquids where the concentration of a certain element follows a normal distribution, and we take a weighted combination of the elements. The concentrations are:

$$C_i \sim \text{IID N}(\mu_i,\sigma_i^2),$$

and the mixture has weights $p_1, ..., p_n$ where $\sum p_i = 1$. I want to find the distribution of the weighted sum $S \equiv \sum p_i C_i$. I have generated values using Monte Carlo simulation, but I would like to know the analytical solution. What is the distribution of this weighted sum for the mixture?

Ben
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RMMA
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    Linear combinations of independent Normal distributions are normal: see https://stats.stackexchange.com/questions/4364. Since Normal distributions are determined by their means and variances, and the mean and variance of your combination is readily calculated, there isn't much to show. – whuber Aug 20 '18 at 21:08
  • @whuber Thanks for your quick response. The mean should be $\sum$$p_i$*$\mu_i$, but how can I calculate the variance? – RMMA Aug 20 '18 at 21:17
  • Its a mixture of normal distributions $\sum_i p_i N(\mu_i,\sigma_i)$. – papgeo Aug 20 '18 at 22:25
  • @papgeo I don't think so. I think it's a linear combination. That becomes clearer when you rewrite a formula from the question as $\sum p_i X_i$ where $X_i \sim \mathcal{N}(\mu_i, \sigma_i).$ – whuber Aug 21 '18 at 00:13
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    @whuber OK, I agree. The question has been edited and very clear now. – papgeo Aug 21 '18 at 07:46
  • It would seem surprising to me to suppose concentrations to be normal. I'd anticipate that in many cases log-concentrations might be better approximated by normal distributions. – Glen_b Aug 21 '18 at 07:51

1 Answers1

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One well-known property of the normal distribution is that a linear combination of independent normal random variables is also a normal random variable. The appropriate mean and variance can be established using ordinary moment rules for linear functions. In this case you have:

$$S \sim \text{N}(\mu_S, \sigma_S^2) \quad \quad \quad \mu_S = \sum p_i \mu_i \quad \quad \quad \sigma_i^2 = \sum p_i^2 \sigma_i^2.$$

Ben
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  • Thanks for your answer. How would a proof look like? – RMMA Aug 21 '18 at 07:26
  • Proof of this property is usually done via characteristic functions, by showing that the product of characteristic functions of normal random variables is still in the normal form. This proof is an implicit use of Fourier transformation to undertake convolutions of random variables. – Ben Aug 21 '18 at 07:44
  • This means if I take C=N(1,$0.4^2$). The result for the linear combination 0.5*C+0.5*C should be C? Using my Monte Carlo approach I get N(1,0.08) as result. – RMMA Aug 21 '18 at 08:51
  • There was an error in the variance formula - now corrected. – Ben Aug 21 '18 at 09:03
  • Thanks! No the result is identical with the result I got using the Monte Carlo method. Still the interpretation is hard for me. That means if I mix two substance which both have a normal distributed concentration of a certain element. The distribution of this element is again normal but the variance is getting smaller. I did expect that 0.5*C+0.5*C = C, where C is a normal distribution. Following your answer this is not the case. – RMMA Aug 21 '18 at 09:12
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    The reason the variance is getting smaller in this case is that you are averaging two independent values. Averages of values are less variable than the individual variables that make up the average. – Ben Aug 21 '18 at 09:40
  • Do you know a book or webpage which shows a proof of the above formula? – RMMA Sep 20 '18 at 13:02