This question is related to another question I asked, but this is more problem-specific.

I have a sequence of A/B currency exchanges for some days. With that data I can calculate the daily returns, and that's what I did. I need to calculate the confidence interval for the expected daily returns of the A/B currency exchange by using the 1.96 rule.

To calculate the confidence interval I would need the standard deviation of the population, but I don't have it. So, I think I need to use the sample standard deviation.

I calculated the sample variance as follows $$SV = \frac{1}{n - 1} \cdot \sum_i ({DR}_{i} - SM)^ 2$$

The sample mean came out to be $0.0083$ and the sample variance $0.6147$. Then I calculated the interval using the formulas

$$\left[L= SM - 1.96 * \sqrt{\frac{SV}{n}}, R= SM + 1.96 * \sqrt{\frac{SV}{n}} \right]$$

where $SM$ and $SV$ are respectively the sample mean and variance I found above. I found that $L = -0.0303$ and $R = 0.0469$.

First of all, is my procedure to find the interval correct? Apart from that, I am still not sure why this procedure would be correct anyway.

What's the exact relation between the number $1.96$ and the $95\%$? And what is this $95\%$ referred exactly to?

If you need the values of the daily returns, I can somehow provide the file.

Thanks for any help!

  • $\begingroup$ Do you know anything about Central Limit Theorem? $\endgroup$ – A.S. Dec 11 '15 at 19:10
  • $\begingroup$ @A.S. Yes, we have covered that already. I know that the CLT has something to do with these topics, but I don't know exactly what. $\endgroup$ – nbro Dec 11 '15 at 19:10
  • $\begingroup$ Did you read my last comment that $SM_n\approx N(\mu,\sigma^2/n)$? $\endgroup$ – A.S. Dec 11 '15 at 19:12
  • $\begingroup$ @A.S. I read your last comment, but I didn't understand well the second part... And yes, I understood what you are saying above, but still I am not sure about what I am doing above... $\endgroup$ – nbro Dec 11 '15 at 19:29

Yes, your method of calculating the confidence interval is correct.

What you calculated as: $$ SE = \sqrt{\frac{SV}{n}}$$ is known as the standard-error of your estimate of the mean. It's basically the standard deviation of your estimator.

Let's assume the true mean were $\mu$. What's referred to the t-stat is: $$ t = \frac{SM - \mu}{SE} $$ follows a t-distribution with $n-1$ degrees of freedom. By the CLT, the distribution of estimator SM converges to the normal distribution but there's a bit of extra complexity for the t-stat here in that we're dividing by the square root of our estimate of the variance, which is a random variable too. The t-distribution is basically the standard normal distribution with somewhat fatter tails.

Where does the 1.96 etc... come from? A $t$ distributed random variable with greater than approximately 1000 degrees of freedom will lie between -1.96 and 1.96 about 95 percent of the time. This can be seen using the inverse of the t-cdf. $F_t^{-1}(.975, 1000) \approx 1.9623 $ and $F_t^{-1}(.025, 1000) \approx -1.9623 $.

  • $\begingroup$ What do you mean by $n -1$ degrees of freedom? $\endgroup$ – nbro Dec 11 '15 at 22:04
  • $\begingroup$ The degrees of freedom is one of the parameters for the t-distribution, it basically describes how many observations were used to estimate the standard error. As the degrees of freedom gets larger, it basically becomes the normal distribution. If you have $n$ independent observations, your data has $n$ degrees of freedom. Your model, (i.e. estimating the 1 variable which is the mean) has $1$ degree of freedom. Loosely speaking, $n-1$ degrees of freedom are leftover to estimate the variance. (This is also why there's an $n-1$ in the variance calculation). And $n-1$ d. of f. are used for t here $\endgroup$ – Matthew Gunn Dec 11 '15 at 22:37
  • $\begingroup$ It's difficult for me to grasp these concepts immediately, because I think my professor has not even talked about the t-distribution, standard-error. $\endgroup$ – nbro Dec 11 '15 at 22:46
  • $\begingroup$ I don't understand though why you say that the standard error is the standard deviation of my estimator. Which estimator? By the way, isn't usually the standard deviation just the square root of the variance, why then we divide by $n$? Sorry, I really new to these things and it's like I don't know well what I am doing. $\endgroup$ – nbro Dec 11 '15 at 22:47
  • $\begingroup$ Let $r$ be a random variable denoting a daily return. There exists some $\mu = E[r]$, that is, the expected value of $r$. Another name is the "population mean." We don't know what that number is! But we'd like an estimate. So we construct an estimator $f(r_1, r_, ..., r_n) = \frac{1}{n} \sum_i r_i$, that is, we use the "sample mean" to estimate the "population mean." If you think about it, you'l see that the sample mean is also a random variable, and hence it has a standard deviation. Just terminology, but for an estimator, the standard dev. is called standard error. $\endgroup$ – Matthew Gunn Dec 11 '15 at 23:20

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