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I have a random variable $z$ for which I've calculated the sample mean $x = \frac{1}{n} \sum z_i$ and the sample standard deviation $s$. How can I calculate the standard deviation of $\frac{1,000,000}{x}$ ?

Edit: More concise and accurate question thanks to Matthew Gunn

Original verbose question:

I have a data set/sample of values and I calculate the average and standard deviation in the end. Now I have a calculated/derived value of this data set/sample that is calculated as $1000000/x$ (with $x$ being the average in the end). Now the question is, what is the standard deviation of my derived value?

I thought I could take the percentage of the standard deviation (standard_dev/average) and just multiple my derived average by it but that doesn't seem right... Is there any relationship between them at all or do I need to compute a new standard deviation by first converting my distribution to my derived/calculated value? (calculate $1000000/x$ for each value and then compute the standard deviation there)

Why do I need this? I want to graph the results with error bars, and I want to graph my derived value as well as the original value.

Further details/domain:

This is benchmarking. I measure run times in μs but want to also display/graph them in iterations per second ("how often could this be run within one second?" which is nice as higher -> better). And of course, I want to show error bars.

Here is an example of values I got (markdown tables don't work here, hope it's good enough):

| Name        | Iterations per Second | Average         | Standard Deviation | Standard Deviation Ratio | Median |
| map.flatten | 1451.6712445317       | 688.8612030905  | 173.9981946453     | 0.2525881758             | 583    |
| flat_map    | 911.7426322002        | 1096.8007469244 | 77.6401735316      | 0.0707878562             | 1056   |
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  • $\begingroup$ This is all rather verbose and somewhat confusing. Is the question essentially, "I have a random variable $z$ for which I've calculated the sample mean $x = \frac{1}{n} \sum z_i $ and the sample standard deviation $s$. How can I calculate the standard deviation of $\frac{1,000,000}{x}$?" $\endgroup$ – Matthew Gunn Jun 11 '16 at 2:39
  • $\begingroup$ Yes I'll add that to the top. Sorry, math has been a while and correct terminology as never been my strong suit so I thought it better to be verbose than leave anything out :) Good summary, thank you. $\endgroup$ – PragTob Jun 11 '16 at 6:19
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A note: reading about Jensen's inequality, $\mathop{\mathbb{E}}\left[1/x \right] \geqslant 1/\mathop{\mathbb{E}}\left[x \right]$ in your case, so I'm not sure it's a useful quantity. Check this thread.

Anyways, you have sample estimates of the mean and standard error of a variable and want to calculate the standard deviation of a transform of that variable.

$$\begin{align}\overline{y}=&\frac{1000000}{\overline{x}}\\SD(\overline y)=&\left|\left|SD(\overline x)\cdot\frac{\partial{\overline y}}{\partial{\overline x}}\right|\right|=\\=&1000000\space\cdot\space \frac{SD(\overline x)}{\overline x ^{2}}\end{align}$$

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    $\begingroup$ I'd drop the word "seems:" there's absolutely no question that $E[1/x] \geq 1 / E[x]$. And I assume you mean $SD(\bar{y})$ and $SD(\bar{x})$ on line two when you apply the delta method? And at the moment, I'm still a bit unclear too on what @PragTob wants... $\endgroup$ – Matthew Gunn Jun 11 '16 at 2:42
  • $\begingroup$ Thank you! And if basic math doesn't fail me, this is the same as $\frac{1000000}{\overline x}\space\cdot\space \frac{SD(\overline x)}{\overline x}$ which is what I did the past years :) Phew, am I glad I don't have to go back and correct all those error bars :) $\endgroup$ – PragTob Jun 11 '16 at 6:39

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