# Variance of resistors in parallel

Suppose you have a set of resistors R, all of which are distributed with mean μ and variance σ.

Consider a section of a circuit with the following layout: (r) || (r+r) || (r+r+r). The equivalent resistance of each part is r, 2r, and 3r. The variance of each section would then be $σ^2$, $2σ^2$, $3σ^2$.

What is the variance in the resistance of the entire circuit?

After sampling several million points, we found that the variance is approximately $.10286\sigma^2$.

How would we arrive to this conclusion analytically?

Edit: Resistance values are assumed to be normally distributed with some mean resistance r and variance $σ^2$.

• I'm not convinced this is an appropriate model to begin with. Are you familiar with the Nyquist-Johnson theory of thermal circuit noise? If you're purposefully doing something different, it would be interesting to see the motivation. Otherwise, it might be worth it to consider a more standard model. :) – cardinal Jul 19 '12 at 2:10
• Yeah, while I was writing my attempt to an answer I also realized that the model apparently is not tractable as it was posed. However, I thought this more like an academic problem rather than a practical one (they are doing simulations, after all). – Néstor Jul 19 '12 at 2:19
• My apologies for having sigma as variance, I originally used VAR and someone edited it to sigma. – lrAndroid Jul 19 '12 at 16:57
• Thanks for the update. I am still interested in the motivation behind this question, if you're willing to add a small bit on that to your question. :) – cardinal Jul 19 '12 at 17:01

The equivalent resistance $R$ of the entire circuit solves $$\frac1R=\sum_{i=1}^{3}\frac{1}{R_i}.$$ One assumes that $R_i=i\mu+\sigma\sqrt{i}Z_i$, for some independent random variables $Z_i$, centered and with variance $1$.

Without further indications, one cannot compute the variance of $R$, hence, to go further, we consider the regime where $$\color{red}{\sigma\ll\mu}.$$ Then, $$\frac{1}{R_i}=\frac1{i\mu}-\frac{\sigma}{\mu^2}\frac{Z_i}{i\sqrt{i}}+\text{higher order terms},$$ hence $$\frac{1}{R}=\frac{a}{\mu}-\frac{\sigma}{\mu^2}Z+\text{higher order terms},$$ where $$a=\sum_{i=1}^{3}\frac1{i}=\frac{11}6,\qquad Z=\sum_{i=1}^{3}\frac{Z_i}{i\sqrt{i}}.$$ One sees that $$\mathrm E(Z)=0,\qquad\mathrm E(Z^2)=b,\qquad b=\sum\limits_{i=1}^{3}\frac1{i^3}=\frac{251}{216}.$$ Furthermore, $$R=\frac{\mu}a-\frac{\sigma}{a^2}Z+\text{higher order terms},$$ Thus, in the limit $\sigma\to0$, $$\mathrm E(R)\approx\frac{\mu}a=\frac6{11}\mu,$$ and $$\text{Var}(R)\approx\sigma^2\cdot\frac{b}{a^4}=\sigma^2\cdot\left(\frac{6}{11}\right)^4\cdot\frac{251}{216}=\sigma^2\cdot0.10286\ldots$$ These asymptotics of $\mathrm E(R)$ and $\text{Var}(R)$ can be generalized to any number of resistances in parallel, each being the result of $n_i$ elementary resistances in series, the elementary resistances being independent and each with mean $\mu$ and variance $\sigma^2$. Then, when $\sigma\to0$, $$\mathrm E(R)\to\frac{\mu}a,\quad \sigma^{-2}\text{Var}(R)\to\frac{b}{a^4},$$ where $$a=\sum_i\frac1{n_i},\quad b=\sum_i\frac1{n_i^3}.$$

I don't think the exact answer depends only on $\mu$ and $\sigma^2$. When you sampled, I suppose you must have used some concrete distribution – probably a normal distribution? In any case, we can calculate the mean and variance of the resistance of the circuit in linear approximation, and then the exact form of the distribution is irrelevant.

The resistance of the circuit is $\left(R_1^{-1}+R_2^{-1}+R_3^{-1}\right)^{-1}$. In linear approximation, the mean and variance of the reciprocal of a random variable with mean $\mu$ and variance $\sigma^2$ are $1/\mu$ and $\sigma^2/\mu^4$, respectively. Thus we have a sum of terms with means $1/\mu$, $1/(2\mu)$ and $1/(3\mu)$ and variances $\sigma^2/\mu^4$, $\sigma^2/(8\mu^4)$ and $\sigma^2/(27\mu^4)$, respectively, which adds up to a mean of $\frac{11}6/\mu$ and a variance of $\frac{251}{216}\sigma^2/\mu^4$. Then taking the reciprocal of that yields a mean of $\frac6{11}\mu$ and a variance of $\left(\frac{251}{216}\sigma^2/\mu^4\right)/\left(\frac{11}6/\mu\right)^4=\frac{1506}{14641}\sigma^2\approx0.10286\sigma^2$, in agreement with your result.

• This is, of course, assuming that the resistors are independent random variables. – Robert Israel Jul 19 '12 at 1:17
• @Robert: Yes (the resistances, rather). That was already assumed in the calculation of the variances $\sigma$, $2\sigma$ and $3\sigma$ in the question, and it makes physical sense (though if we take all the resistors from the same batch of production, their resistances will be somewhat correlated). – joriki Jul 19 '12 at 1:20
• In a real design, of course, the resistances are far from independent rvs. In fact, much work goes into layout to make some groups of elements track each other (called ''matching', unsurprisingly). – copper.hat Jul 19 '12 at 6:48
• Are you using $\sigma = E (X-EX)^2$? I am more used to seeing this written as $\sigma^2$. – copper.hat Jul 19 '12 at 7:00
• @copper.hat: You're quite right about $\sigma^2$, of course -- I'd adopted the notation used in the question without thinking. – joriki Jul 19 '12 at 7:05

This depends on the shape of the distribution for the resistance. Without knowing the distribution, I can't even say the average resistance, although I think there are constraints.

So, let's pick a distribution which is tractible: Let $s$ be the standard deviation of the resistance of one resistor. Let the resistance be $\mu \pm s$, with each sign occurring with with probability $1/2$. This gives us $2^6 = 64$ cases to consider, or $2 \times 3 \times 4=24$ if we combine some cases. Of course we'll assume the resistances are independent.

If we choose $\mu = 100$ and $s=1$ then the mean is $54.543291$ (slightly lower than $100 \times \frac{6}{11}$), and the variance is $0.102864$. If we choose $\mu = 5$ and $s=1$, then the variance is $0.103693$.

Here is a power series expansion for the ratios between the variances when the mean is $1$ and the variance is $x$: $\frac{1506}{14641} + \frac{36000}{1771561}x + \frac{218016}{19487171}x^2 + O(x^3)$. When $x$ is small, the dominant term is $\frac{1506}{14641} = 0.102862$.

While the question you ask technically depends on the distribution, you are probably interested in situations where the standard deviation is small compared with the mean, and I think there is a well-defined limit which doesn't depend on the distribution. Linearize the dependence of the circuit's resistance as a function of the resistances of each piece:

$$C = \frac {1}{1/R_1 + 1/(R_2+R_3) + 1/(R_4 + R_5 + R_6)}$$

$$\approx \frac{6}{11} \mu + \sum_{i=1}^6 (R_i - \mu)\frac {\partial C}{\partial R_i}(\mu,\mu,\mu,\mu,\mu,\mu)$$

$$\therefore \text{Var}(C) \approx \sum_{i=1}^6 \text{Var}(R_i)\bigg(\frac {\partial C}{\partial R_i}(\mu,\mu,\mu,\mu,\mu,\mu)\bigg)^2$$

With this specific circuit, the scaled partial derivatives are $\frac {36}{121}, \frac{9}{121} ,\frac{9}{121}, \frac{4}{121},\frac{4}{121},\frac{4}{121}$, and

$$\bigg(\frac{36}{121}\bigg)^2 + 2\bigg(\frac{9}{121}\bigg)^2 + 3\bigg(\frac{4}{121}\bigg)^2 = \frac{1506}{14641} = 0.102862$$

• This reminds me of Multivariate delta theorem, i.e. $R_1,R_2,R_3$ has mean $\mu,2\mu,3\mu$ and variance $\sigma^2, 2\sigma^2, 3\sigma^2$ respectively, then $g(R_1,R_2,R_3) = ((1/R_1)+(1/R_2)+(1/R_3))^{-1}$ should have asymptotic variance as $\nabla g(\mu)\Sigma\nabla g(\mu)'$, where $\nabla g(\mu) = (\frac{36}{121},\frac{9}{121},\frac{4}{121})$ and $\Sigma = $\left( \begin{array}{ccc}. \sigma^2 & 0 & 0 \\ 0 & 2\sigma^2 & 0 \\ 0 & 0 & 3\sigma^2 \end{array} \right)$$. The final answer is same as @Douglas Zare and OP, that is 0.1028$\sigma^2$. – VitalStatistix Jul 19 '12 at 4:27

I warn that, as I reasoned it, this is a long answer, but maybe someone can come up with something better starting from my attempt (which may not be optimal). Also, I misread the original OPs question and thought it said that the resistances where normally distributed. I'll leave the answer anyways, but that's an underlying supposition.

1. Physical reasoning of the problem

My reasoning is as follows: recall that, for resistors that are in paralel, the equivalent resistance $R_{eq}$ is given by:

$$R_{eq}^{-1}=\sum_{i}^{N}\frac{1}{R_i},$$

where $R_i$ are the resistances of each part of the circuit. In your case, this gives us

$$R_{eq}=\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)^{-1},\ \ \ (*)$$ where $R_1$ is the part of the circuit with 1 resistance, and has therefore a normal distribution with mean $\mu$ and variance $\sigma^2$, and by the same reasoning $R_2\sim N(2\mu,2\sigma^2)$ is the equivalent resistance of the part of the circuit with two resistances and, finally, $R_3\sim N(3\mu,3\sigma^2)$ is the equivalent resistance of the part of the circuit with three resistances. You ought to find the distribution of $R_{eq}$ and from there obtain the variance of it.

2. Obtaining the distribution of $R_{eq}$

One way to find the distribution is by noting that: $$p(R_{eq})=\int p(R_{eq},R_1,R_2,R_3)dR_1dR_2dR_3=\int p(R_1|R_{eq},R_2,R_3)p(R_{eq},R_2,R_3)dR_1dR_2dR_3.\ \ \ (1)$$ From here, we also note that we can write $$p(R_{eq},R_2,R_3)=p(R_2|R_{eq},R_3)p(R_{eq},R_3)=p(R_2|R_{eq},R_3)p(R_{eq}|R_3)p(R_3)$$ (which was obtained via the Bayes Theorem), which, assuming independance between $R_1$, $R_2$ and $R_3$ (which is physically plausible), can be written as $$p(R_{eq},R_2,R_3)=p(R_2|R_{eq})p(R_{eq}|R_3)p(R_3).$$ Replacing this in $(1)$ and noting that another consequence of the independance between the three resistances is that $p(R_1|R_{eq},R_2,R_3)=p(R_1|R_{eq})$, we get: $$p(R_{eq})=\int p(R_1|R_{eq})p(R_2|R_{eq})p(R_{eq}|R_3)p(R_3)dR_1dR_2dR_3=\int p(R_{eq}|R_3)p(R_3)dR_3.\ \ \ (2)$$ Our last problem then is to find $p(R_{eq}|R_3)$, i.e., the distribution of the r.v. $R_{eq}|R_3$. This problem is analogous to the one we found here, except that now you replace $R_3$ in eq. $(*)$ by a constant, say, $r_3$. Following the same arguments as above, you can find that $$p(R_{eq}|R_3)=\int p(R_{eq}|R_2,R_3)p(R_2)dR_2.\ \ \ (3)$$ Apparently the rest is replacing the known distributions, except for a little problem: the distribution of $R_{eq}|R_2,R_3$ can be obtained from $(*)$ by noting that $X_1$ is gaussian, so, you essentially need to find the distribution of the random variable $$W=\left(\frac{1}{X}+a+b\right)^{-1},$$ where $a$ and $b$ are constants, and $X$ is gaussian with mean $\mu$ and variance $\sigma^2$. If my calculations are correct, this distribution is: $$p(W)=\frac{1}{[1-W(a+b)]^2}\frac{1}{\sqrt{2\pi \sigma^2}}\text{exp}\left(-\frac{X(W)-\mu}{2\sigma^2}\right),$$ where, $$X(W)=\frac{1}{W^{-1}-a-b},$$ so $R_{eq}|R_2,R_3$'s distribution would be $$p(R_{eq}|R_2,R_3)=\frac{1}{[1-R_{eq}(a+b)]^2}\frac{1}{\sqrt{2\pi \sigma^2}}\text{exp}\left(-\frac{X(R_{eq})-\mu}{2\sigma^2}\right),$$ where $a=1/R_2$ and $b=1/R_3$. The thing is that I don't know if this is analytically tractable in order to solve the integral in equation $(3)$, which then will lead us to solve the poblem by replacing it's result in equation $(2)$. At least to me at this time of the night it is not.

• You are assuming a normal distribution, even though resistance can't be negative? My guess is that this will make the variance of the circuit diverge. – Douglas Zare Jul 19 '12 at 2:18
• I know, that bottered me too, but in practice it really depends on the values of $\mu$ and $\sigma^2$. If $\mu>>0$ and $\mu>>\sigma$, then we can "save" the model. In normal conditions, the dispersion of a resistance is not very high, so the last assumption is clearly met. This was something that initially bothered me too when people modeled height as a normal random variable, but by the same reason that I gave here, some people here at Stack-exchange made me feel ok with it :-). – Néstor Jul 19 '12 at 2:24
• Hmm, I think modeling height as normal is so bad that I use it as an example of a distribution which obviously isn't normal. I suppose it might not be terrible if you have a population of healthy adult men from the same genetic background. However, I'd want to hear from a biologist that this is ok. The reasoning I have too often heard that each bone's size is independent is total nonsense. – Douglas Zare Jul 19 '12 at 2:37
• I just realized that the resistances weren't normally distributed (I could swear that I read they where on the original OPs answer, but I think it was just my imagination). – Néstor Jul 19 '12 at 3:48