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Say I simulate from a normal distribution $N(\mu, \sigma^2)$ 10.000 times using a couple of different methods.

I can calculate the standard error as $s/\sqrt{10000}$ where $s$ is the sample standard deviation.

What I don't understand is .... $s$ is the standard deviation of the sample. The sample are generated by simulating a normal distribution. Hence, by that very fact, the sample standard deviation of all simulations will be equal to $\sigma$.

So, for all my simulation methods, the standard error will be the same. It will be roughly $\sigma /\sqrt{10000}$. So what is the point of calculating it? It is already known before-hand and is constant for all my simulations, as long as I am simulating the same distribution?

So what is the point of calculating this number? Let me give you an example.

Say you write in R:

sample_1 <- rnorm(10000, mu, sigma)

u <- runif(10000)
sample_2 <- qnorm(u)

Now, both sample_1 and sample_2 are 10.000 simulations from a $N(\mu, \sigma^2)$ distribution... In either case, the standard error will roughly be $\sigma/\sqrt{10.000}$. This number seems pointless to me. It tells me nothing about the relative performance of $sample_1$ vs $sample_2$ ....the standard error is the same?

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    $\begingroup$ In many cases you have the sample but never know $\mu$ or $\sigma$. You then use $\bar x$ to estimate $\mu$, and use $s / \sqrt{n}$ to estimate the magnitude of the possible error in making that estimate of $\mu$ $\endgroup$
    – Henry
    Commented Sep 22, 2019 at 10:21
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    $\begingroup$ @Henry , I think that's the answer. You might convert that comment to an answer. $\endgroup$
    – Sal Mangiafico
    Commented Sep 22, 2019 at 12:48

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As requested in comments:

In many cases you have the sample but never know $\mu$ or $\sigma$.

You then use ${\bar x}$ to estimate $\mu$, and use $s / \sqrt{n}$ to estimate the magnitude of the possible error in making that estimate of $\mu$

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  • $\begingroup$ But why use $s$ when you know $\sigma$? $\endgroup$
    – Dave
    Commented Sep 22, 2019 at 14:55
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    $\begingroup$ @Dave If you know $\mu$ then this is all irrelevant. If you know $\sigma$ but not $\mu$ (how?) then use $\bar x$ and then $\sigma / \sqrt{n}$ for the standard error. If you know neither $\mu$ nor $\sigma$ then use $\bar x$ and then $s / \sqrt{n}$ for the estimate of the standard error $\endgroup$
    – Henry
    Commented Sep 22, 2019 at 16:25
  • $\begingroup$ Z-test has known $\sigma$ and unknown $\mu$. $\endgroup$
    – Dave
    Commented Sep 22, 2019 at 16:30

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