Determine a product’s rating given a known 3-sigma tolerance

distributionsempirical-cumulative-distr-fnprobabilitysigma-algebra

Consider a 100Ω resistor with a 10% tolerance. We can assume this is the 3-sigma value since this is typical in manufacturing. Thus, we can expect ≈99.73% of such resistors to range from 90Ω to 110Ω.

Putting aside the fact that the manufacturer will re-bin resistors with a 5% tolerance, I want to figure out the average resistor's Ohms value and express this as a rating.

One idea I had was to use rms. Something like this (using ≈1.11 to convert rms to average):

$\delta=x±10\%=100±10\%=10$

$r=\frac{\delta}{\sqrt{2}\times1.11}+(x-\delta)=\frac{10}{\sqrt{2}\times1.11}+(100-10)\approx96.37\%$

Remember, I want a rating, hence the percentage. Using this same approach, a 100Ω resistor with a 5% tolerance would have a rating of ≈98.19%.

How accurate is this approach mathematically?

Alternate approach:

Looking at the normal distribution, I realize that integration will answer my question.

I know the height of the mode in a normal density is $\frac{1}{\sqrt{2\pi\sigma}}$.

What I don't know is the height at 3-sigma, or how to use integration to solve this question.

I'm a software engineer, but not a statistician. How do I calculate this?

Best Answer

First, consider that we don't know what is the true distribution of resistance (you might want to look here). But, for simplicity (and to go well with the $3\sigma$ assumption), let's say it has a normal distribution. That is, if we have a $100\Omega$ resistor with a silver tolerance band (10%), then $P(R\in[90\Omega,110\Omega])=0.997$. That is the meaning of confidence intervals and the $3\sigma$ notation: It means that no more than 3 out of 1000 resistors produced in this manufacturing line had resistance of more than $110\Omega$ or less than $90\Omega$.

Now, we can decompose this CI in order to get an estimate of the average resistance, but spoiler alert: we'll get that the average is exactly $100\Omega$.

The middle value of the CI is obviously $100\Omega$. We assume $3\sigma=10\Omega$ so the SD for a resistor from this production line is $3.33\Omega$. You can try finding which value of $\mu$ would minimize the equation $ \sqrt{ \int_{-\infty}^{+\infty} (x-\mu)^2 f(x)dx } - \sigma$ (where $f(x)$ is the normal PDF with parameters $\mu,\sigma$) but then again you'll get that the minimizer is $\mu=100\Omega$.

Should you find any alternative distribution for the resistance of a resistor (I mean, other than the normal), we can try constructing a CI fitting it using the delta method.

Regarding the RMS approach - As far as my knowledge goes (in both electronics and stats), we use RMS to get an approximation for a bunch of measurements, whether it's a dataset or the response to a wave input. I don't think it belongs here, and if you insist on using it then keep in mind that taking $(x-\delta)$ alone is not enough, you should also take $(x+\delta)$ into calculation.

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Can any Probability Distribution be Integrated

There are four little errors.

In computing the denominator you take a power of 4 (instead of 3)
You had a wrong mean vector x_two2 = y - 1
You integrate over the joint distribution f(x,y,z) instead of the conditional distribution f(x,y,z)/f(x)

In the corrected code below it is simply dividing by dnorm(1) which can be done because x is independent from y and z.
The range of integration is small (the runif functions)

#define constants needed for the multivariate normal
    sigma1 <- c(1,0,0,0,1,0,0,0,1)
      sigma <- matrix(sigma1, nrow=3, ncol= 3, byrow = TRUE)
      sigma_inv <- solve(sigma)
      sigma_det <- det(sigma)
      denom = sqrt( (2*pi)^3 * sigma_det)

#mixture model
    target <- function(y,z)
      
    {
      x_one = 2 - 1
      x_two = y - 1
      x_three = z - 1
     
        
      x_t = c(x_one, x_two, x_three)
      x_t_one <- matrix(x_t, nrow=3, ncol= 1, byrow = TRUE)
      x_t_two =  matrix(x_t, nrow=1, ncol= 3, byrow = TRUE)
      
      
     
      num = exp(-0.5 * x_t_two  %*%  sigma_inv  %*%  x_t_one)
        
      answer_1 = num/denom
 
    
      x_one2 = 2 - 1
      x_two2 = y - 0
      x_three2 = z - 1
     
        
      x_t2 = c(x_one2, x_two2, x_three2)
      x_t_one2 <- matrix(x_t2, nrow=3, ncol= 1, byrow = TRUE)
      x_t_two2 =  matrix(x_t2, nrow=1, ncol= 3, byrow = TRUE)
      
      
      # In this part, as it's (x-mu)^T * SIGMA * (x-mu)
      
   
      num2 = exp(-0.5 * x_t_two2  %*%  sigma_inv  %*%  x_t_one2)
        
      answer_2 = num2/denom
      return(0.3*answer_1 + 0.7*answer_2)
    }

means_vec<-c()
for(j in 1:10){
    set.seed(j)
    box_length = 10
    u_y<-runif(10000,-box_length/2,box_length/2)
    u_z<-runif(10000,-box_length/2,box_length/2)
    sim_vec<-c()
    for(i in 1:length(u_y)){
        sim_vec[i]<-target(y=u_y[i],z=u_z[i])
    }
    means_vec[j]<-mean(sim_vec)
}
print(means_vec*box_length^2/dnorm(1))

mean(means_vec)*box_length^2/dnorm(1)

PDF of $Z=X^2 + Y^2$ where $X,Y\sim N(0,\sigma)$

Following the comment by whuber, for problems like this that involve convolutions of IID random variables, it is generally simpler to work with the characteristic function than with the density function. Using the law of the unconscious statistician we can obtain the characteristic function for $X^2$ as follows:

$$\begin{align} \phi_{X^2}(t) &\equiv \mathbb{E}(\exp(itX^2)) \\[16pt] &= \int \limits_{-\infty}^\infty \exp(it x^2) \cdot \text{N}(x|0, \sigma^2) \ dx \\[6pt] &= \int \limits_{-\infty}^\infty \exp(it x^2) \cdot \frac{1}{\sqrt{2 \pi \sigma^2}} \cdot \exp \bigg( -\frac{1}{2 \sigma^2} \cdot x^2 \bigg) \ dx \\[6pt] &= \int \limits_{-\infty}^\infty \frac{1}{\sqrt{2 \pi \sigma^2}} \cdot \exp \bigg( -\frac{1-2it \sigma^2}{2 \sigma^2} \cdot x^2 \bigg) \ dx \\[6pt] &= \frac{1}{\sqrt{1-2it \sigma^2}} \int \limits_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} \sqrt{\frac{1-2it \sigma^2}{\sigma^2}} \cdot \exp \bigg( -\frac{1-2it \sigma^2}{2 \sigma^2} \cdot x^2 \bigg) \ dx \\[6pt] &= \frac{1}{\sqrt{1-2it \sigma^2}} \int \limits_{-\infty}^\infty \text{N}\bigg( x \bigg| 0, \frac{\sigma^2}{1-2it \sigma^2} \bigg) \ dx \\[6pt] &= \frac{1}{\sqrt{1-2it \sigma^2}}. \\[6pt] \end{align}$$

(And of course, we have $\phi_{X^2}(t) = \phi_{Y^2}(t)$ in this case so the latter characteristic function is the same.) We then have:

$$\begin{align} \phi_Z(t) &\equiv \mathbb{E}(\exp(itZ)) \\[16pt] &= \mathbb{E}(\exp(itX^2 + itY^2)) \\[16pt] &= \mathbb{E}(\exp(itX^2)) \cdot \mathbb{E}(\exp(itY^2)) \\[12pt] &= \frac{1}{\sqrt{1-2it \sigma^2}} \cdot \frac{1}{\sqrt{1-2it \sigma^2}} \\[6pt] &= \frac{1}{1-2it \sigma^2}. \\[6pt] \end{align}$$

This is the characteristic function for the scaled chi-squared distribution with two degrees-of-freedom. Using the fact that the characteristic function is a unique representative of the distribution, you then have:

$$Z \sim \sigma^2 \cdot \text{ChiSq}(\text{df} = 2).$$

Best Answer

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Can any Probability Distribution be Integrated

PDF of $Z=X^2 + Y^2$ where $X,Y\sim N(0,\sigma)$

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