# Calculating Variance of Eq. with random variables

• shakystew
In summary, the user is attempting to calculate heat transfer across a medium with known material properties. They are missing one variable and are unsure how to calculate it. They have estimated the value using engineering parameters and have used a random number generator with 10% perturbations to get 1E9 histories. They are now wondering how to calculate the variance and if it is possible to create a PDF for the variable. They have received advice to use a triangular distribution for the variable.
shakystew

## Homework Statement

I am attempting to calculate a heat transfer across a medium with known material properties. I have the equation and all but one variable I have an exact answer for. I require the variance of my answer.

## Homework Equations

I know ALL variables (ie numerical value) except the the last h. So far, using some engineering parameters for the problem, I have estimated the value by taking 10% perturbations of the normal value of h (that is, the value is usually ~3.0 for this problem, and I have used logic to say that with the additional medium, the value will only affect AT GREATEST +/- 10%).

I am unsure on how to actually calculate the value now. I have ran the equation in MATLAB using random number generator from 2.7-3.3 (10% perturbations) with 1E9 histories. Can I use this value to get the variance? (The output is not normal/gaussian).

Last edited:
Assuming everything except ##h## is constant, the variance is simply
$$\text{var}\Delta T = \left(\frac{q'}{2\pi R_{00}}\right)^2 \text{var}\left(\frac{1}{h}\right)$$
This follows from the following basic property of variance: if ##x## is a random variable and ##a## and ##b## are constants, then ##\text{var}(a(x+b)) = a^2 \text{var}(x)##. In other words, the constant offset ##b## does not affect the variance, and the multiplicative scale factor ##a## multiplies the variance by ##a^2##.

There is no general property relating ##\text{var}(1/h)## to ##\text{var}(h)##. How they relate depends on how ##h## is distributed. You can start with the definition of variance:

$$\text{var}\left(\frac{1}{h}\right) = E\left[\left(\frac{1}{h}\right)^2\right] - \left(E\left[\frac{1}{h}\right]\right)^2$$

where the expected values are calculated in the usual way using the integral definition. It looks like you are assuming that ##h## is uniformly distributed in the interval ##[2.7, 3.3]##, so you should be able to get a closed form answer without resorting to numerical approximation.

1 person
Creating a PDF for such variable

Thank you jbunniii for the quick response!

If I didn't know the value of h, how could I make a PDF for such variable? Wouldn't that require an equation of h to find such parameters? Or, if I do use those contraints from |2.7 , 3.3|, how would I go about making a PFD?

My advisor said I should first make a triangular distribution of the variable. To use a triangular distribution, I must know the aand b (where the value is zero) as well as the peak value, c.

shakystew said:
Thank you jbunniii for the quick response!

If I didn't know the value of h, how could I make a PDF for such variable? Wouldn't that require an equation of h to find such parameters? Or, if I do use those contraints from |2.7 , 3.3|, how would I go about making a PFD?
There's no general answer to this - it depends on your specific problem. Where are the values of ##h## coming from? Do you have some measurement data? If so, you can try fitting a distribution to the data. [Sorry, I'm not an expert regarding how to do that, but I know there are statistical methods for doing this.]

Or maybe you just know that the value cannot be smaller than ##2.7## or larger than ##3.3##, so the distribution must be constrained to that interval. If you know nothing else, a uniform distribution in that interval may be reasonable. If you think the average ##3.0##is more likely than ##2.7## or ##3.3##, then a triangle or some other "peaked" distribution may be reasonable.

If you only know a mean and a variance for ##h## then a normal/gaussian assumption may be reasonable. It all depends on what you know about ##h##.

## 1. What is the formula for calculating variance of a set of random variables?

The formula for calculating variance is:
Var(X) = E[(X - µ)^2]
where X is the random variable and µ is the mean of the set of values.

## 2. Why is calculating variance important in statistical analysis?

Calculating variance allows us to measure the spread or variability of a set of data. It helps us understand the distribution of the data and make comparisons between different sets of data. It is also a fundamental concept in probability and statistics.

## 3. What is the difference between population variance and sample variance?

Population variance is calculated using the entire population of data, while sample variance is calculated using a smaller subset of the population. Sample variance is an estimate of the population variance, and it is used when the entire population cannot be measured or is too large to calculate.

## 4. Can variance be negative?

No, variance cannot be negative. It is always a non-negative value because it is calculated by squaring the differences between the data points and the mean, which eliminates the possibility of negative values.

## 5. How can we interpret the variance of a set of random variables?

The variance of a set of random variables measures the average distance of the data points from the mean. A smaller variance indicates that the data points are closer to the mean, and a larger variance indicates that the data points are more spread out from the mean. It is also useful in identifying outliers or extreme values in the data.

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