I am trying to work some things in R and I am having trouble understanding some of the instructions.

I generated $1000$ samples of size $5$ from the standard normal distribution, and I calculated the mean of the sample variance of these. Now I want to know what the sample variance of my sample of sample variances is. But I am not sure I understand really what this means, nor how to implement this in R.

Further, I am asked to overlay the histogram I generated from my sample with a histogram of the theoretical density of the sampling distribution. What does this mean? Ie, what is meant by the theoretical density of the sampling distribution of the sample variance.

I know all my samples are coming from standard normal, where $\sigma^{2}=1$

and I know that if $X_{N}=X_{1}+...+X_{1000}$ would be $N(0,\frac{\sigma^{2}}{1000})$, is this at all what is being referred to?

I will appreciate any help and advice. Thank you


The distribution of the sample variance $S^2$ is given by $(n-1)S^2/\sigma^2 \sim \chi^2(n-1)$. I'm guessing that you are asked to provide an illustration of this relationship using R. Consider the following simulation.

 m = 1000;  n = 5;  x = rnorm(m*n)
 DTA = matrix(x, nrow=m)  # each row a sample of size n
 v = apply(DTA, 1, var)   # sample variances of m rows
 hist((n-1)*v, prob=T, col="wheat", ylim=c(0,.2))
   curve(dchisq(x, n-1), lwd=2, col="blue", add=T)
   lines(density((n-1)*v), lwd=2, col="darkgreen")
 ## 1.003081 
 ## 0.4881987

enter image description here

This may not be exactly what you are being asked for, but it may point you in the right direction. I have overlaid a density curve on the histogram. I'm not sure what kind of histogram could be superimposed.

Probably, an important message here is that the relevant chi-squared distribution has df = n-1, not df = n. You can try superimposing the density of $Chisq(5)$ and you'll see it doesn't fit the histogram at all well.

$Addendum:$ I don't know if you know about density estimators, but for good measure, I also superimposed a density estimator (smoothed histogram) in green. For this particular simulation run the theoretical curve and the density estimator agree pretty well, but if you run the program several times you will get some cases in which the agreement isn't so good. (If you use m = 10,000, results will be more stable.)

Please let me know if you can make sense of this to finish your project. What is the variance of $Chisq(4)$? If you don't know, look at the Wikipedia article on 'Chi-squared distribution'.

Addendum per Comment from @Quality: Because $(n-1)S^2/\sigma^2 \sim Chisq(4),$ we have $V[4S^2] = 2(4)$ or $V(S^2) = 8/16 = 1/2$. Also, v in the program represents $S^2,$ so it is not surprising that var(v) returns $0.488 \approx 0.5$ within simulation error. (Because variances are on a squared-unit scale, the margin of simulation error is numerically larger for variances than for means: Several additional runs of the program gave values between 0.47 and 0.59. Use m=10^6 for a slower run with better accuracy.)

  • $\begingroup$ Thank you so far, yes, as of now I had calculated sample.var(n=5,N=1000) and a mean(sample.var(n=5,N=1000)) just to see what was going on. Also, yes I am familiar with chi squared and its relation to what you had above.So for some clarifications, the yellow bricks represent the results from the samples, while the lines are from the chi squared? Very much appreciated. $\endgroup$ – Quality Feb 1 '16 at 6:11
  • $\begingroup$ can you help explain this? For example, when I search it up I see that the expected value of chi squared with n deg of freedom is n, and variance 2n, so wouldnt this seemigly imply that in my example above the expected value was 4? $\endgroup$ – Quality Feb 2 '16 at 20:36
  • $\begingroup$ @Quality: I put another 'Addendum' at the end of my Answer, which I hope helps resolve this issue. $\endgroup$ – BruceET Feb 2 '16 at 22:30

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