# different types of Averages : prove there is an inequality between different averages

This is something I have been curious based on observations I have made by analyzing data from the energy industry. The context is you have monthly data for energy used per month and the cost per month.

So for example lets say that we have twelve months of data(I have a spreadsheet attached with sample). For every month we can divide the cost by usage, and we get a pricing or also referred to as a rate. If we average these 12 rates we get one annual rate.

That is method number 1 to create an annual rate. Another way is to add all the energy usage for all 12 months, then add all cost, and divide the two to get another kind of average or rate.![Average Methods formula]

Note in the image: $c$ represents cost data and $m$ is usage data. $$\left(\sum_i C_i\right) / \left(\sum_i M_i\right) == \frac1N \sum_i(C_i/M_i)$$ So i put a bunch of equal signs in between, I want to find out if there exists an inequality between these 2 averages or rates.

I am not sure how one can prove this,not sure what techniques. Either Induction or using somekind of triangle inequality. I am just wondering if someone can figure out the approach. Also when I do a numerical example, with data, it seems that averaging monthly rates is greater than getting an average of total sum divided by total sum.

Looks like the message I am getting from this app is that I can`t post images until I get up to 10 posts.Also I can't see where I can attach files like spreadsheet. Also I don't know how to make complicated formulas like making summation symbols here.

Hope someone can give me some pointers on how to create mathematical formulas. P

• For writing $\LaTeX$ you could start here Jun 26, 2013 at 2:14
• Hi,I guess that my way of writing it was edited by someone else. Specifically by Omnomnomnom. Thanks who ever you are, you did a good job. P
– Palu
Jun 26, 2013 at 3:36

On the left side, were the sum of the costs gets divided by the sum of the usages, if the cost and usage are very high one month they will dominate the average, while on the right side they will have no more impact than any other month. Let's pick two examples. The first,$$\begin {array} {r c c} month&C_i&M_i\\1&100&100\\2&2&1 \end {array}$$ Then $\left(\sum_i C_i\right) / \left(\sum_i M_i\right) =\frac {102}{101} \approx 1\\\frac1N \sum_i(C_i/M_i)=\frac 32$
The second $$\begin {array} {r c c} month&C_i&M_i\\1&100&100\\2&1&2 \end {array}$$ Then $\left(\sum_i C_i\right) / \left(\sum_i M_i\right) =\frac {101}{102} \approx 1\\\frac1N \sum_i(C_i/M_i)=\frac 34$
So it can go either way.

• OK, I see you have done something. This does seem to be an extreme case. i will need to reformulate my idea possibly. Thanks for illustrating this extreme case Ross. Palu
– Palu
Jun 26, 2013 at 3:53
• OK, I now realize looking at my data, that there is a relationship between the Cost and Usage, usually the Cost is in the range of say 8% to 15% of the Usage, always, this is for electricity usage and for nat.gas it is in the proportion cost is 20 to 40% of the usage. So I have been looking at data that always maintains this kind of proportion or ratio.
– Palu
Jun 26, 2013 at 4:01
• The exact percentages are not important, as you can change units and change the percentages (measure cost in cents instead of dollars). What matters is how the volume varies. If the usage is the same every month, the two averages will be equal. If the usage is high one month, that month counts much higher on the left than on the right. That is because the sums for the left ratio will be dominated by that month, which was the point of my example. Jun 26, 2013 at 4:06
• Even if the variation is smaller, the basic truth will not change, it just won't be so evident. Jun 26, 2013 at 4:07
• ok, that is interesting. I still have to convince myself if that is the case. I see based on you switching month number2, where you put the usage to be value 2 and cost to be value 1. it caused a change. But that is the thing based on the ratio the cost will always be lower than the usage.So your second case represents the situation that I would be encountering.
– Palu
Jun 26, 2013 at 4:15