Meaning of "sample size" when sample unit is ambiguous

Question

Suppose I am collecting data on how much money is processed by the 16 banks in an economy. I want to quantify how "concentrated" the flow of money is—that is, the extent to which the larger banks process more money than the smaller banks.

In a certain random interval of time, I recorded the transaction volume in dollars and as a percentage at each bank. To avoid dependency issues, let's assume that the ranks of the banks are known prior to sampling. The banks are then referred to by these fixed ranks; therefore, in this random sample, the transaction volume does not strictly decrease as rank increases.

bank    volume        pc    
1        75800  0.336440
2        49500  0.219707
3        50200  0.222814
4         7900  0.035064
5         9000  0.039947
6         6200  0.027519
7         6500  0.028850
8         5300  0.023524
9         4800  0.021305
10        4100  0.018198
11        1100  0.004882
12        1100  0.004882
13        1100  0.004882
14         800  0.003551
15        1400  0.006214
16         500  0.002219

The first and third columns then describe a histogram that estimates the pmf $f(x)$, where $X$ is a random variable denoting the rank of the bank at which a given dollar is transacted.

To estimate the sample mean, or the expected rank of the bank at which a random dollar is transacted, I take the dot product of bank and pc, which comes out to 3.0203. This is easy, and we don't need any corrections that depend on the sample size.

But I am interested in computing unbiased estimators for statistics based on higher moments. In the formula for the sample variance (for example), to get an unbiased estimator we correct the maximum likelihood estimator by multiplying it by $\frac{N}{N-1}$:

$$s^2 = \frac{1}{N-1} \sum_i^N (x_i - \bar x)^2$$

I'm not sure what number I should use for $N$ in a context like this. One option could be to treat each dollar as an $x_i$ (a trial). Then our (enormous) "sample" would consist of 75800 1s, 50500 2s, and so on, and we can use the correction above with $\frac{N}{N-1} = \frac{225300}{225299} \approx 1$. But you could just as easily argue for using hundreds of dollars, or cents, or whatever as the trial unit, which would drastically alter the sample size.

• What does sample size mean in a situation like this?
• What "sample" should I use to compute unbiased estimators for the standard deviation, skewness, and kurtosis of this distribution?

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I realize there are other tools economists use to quantify how concentrated the flow of money is. I am not asking about these other tools, but rather seeking a conceptual understanding of sample size as it relates to unbiased estimators.

Answer 1

There are some problems with your approach:

• Mean of ranks (ordinal data), may or may not make sense, depending on your application. Is it sensible to treat a bank of rank 15 as 15 times a bank of rank 1? I.e. is the mean between a bank of rank 1 and one of rank 15 (given equal share) a bank of rank 8? You should think about if your estimate makes sense at all, or if it is better to use some other statistic (e.g. the median).
• Be very clear about the question you are asking. "What is the rank of the bank, at which any random dollar ist most likely to be processed". This question is easy to pose and has a straight forward answer: The bank with the highest volume (no need to take the mean). This may or may not be what you need, but you should take care that your estimators meet the question you are asking.

To your actual question: Have a look at how you calculated the mean, and which numbers went into there. You said you did not need the sample size to calculate the mean, but you fooled yourself there. Mean always requires the sample size, but you just used a calculation that hid where it was used.

Have a look at how you calculated the pc values (if I understood this correctly).

$pc_i=\frac{amount_i}{\sum_{i=1}^N amount_i}$.

You might already see where the sample size comes in here. If not let me re-formulate the dot product:

$E[rank]=\sum_{i=1}^N rank_i \cdot pc_i=\sum_{i=1}^N rank_i \frac{amount_i}{\sum_{i=1}^N amount_i}=\frac{\sum_{i=1}^N rank_i\cdot amount_i}{\sum_{i=1}^N amount_i}$

Think about this for a while, until you see where the sample size appears.

Further points:

• You said that your sample size changes if you take cents for example. With the above example, you can see that this is absolutely correct. Your numbers get a different meaning if you change from dollars to cents. But at the same time, your $x_i$ also change (note that $N$ also appears in the sum, so the number of $x_i$ also changes). In most cases, this should cancel out. If you believe your statistics don't make sense after you change the unit, then you may not have the right estimator to answer your question.
• The correction $N-1$ for the variance is used when estimating the variance of a normally distributed variable. With normally distributed variables, changing the unit in a way as you do here should cancel out (not tested, but my intuition says it should). If it doesn't you may use a correction for a normally distributed variable, that is not normally distributed.

Answer 2

Suppose I am collecting data on how much money is processed by the 16 banks in an economy. [...]

So your units of observations are the banks, you are looking at a sample of $N=16$ banks. If you would somehow take random sample of the dollar bills and tracked what happens with each of them, then the units of observations would be dollars.

However, by using such "unbiased" computation you would totally diverge from your intended computation that considered the ranks etc. If for computing the intended quantity you are using some kind of weighted estimator, you probably should use weighted variance as well, for the same reasons.

Answer 3

Asking 'what is my sample size' can be restated as 'how many independent realizations of my data generating process have there been'. Since your data was generated by transations arriving at banks I would see the individual transactions as the unit of observation, and consider my sample size as the number of transactions that led to the data that I have.

I'm not sure what you are really estimating when you want the 'standard deviation of the rank at which a random dollar is transacted'. But the variance of this rank surely depends on the degree to which dollars arrive independently of each other, and your dataset as presented does not include this information. You could probably learn more by bootstrapping with different assumptions re transaction size.