1
$\begingroup$

I am trying to think about the distribution of a random process. Here's how you would generate the sequence: for each sample of size k (sampled from i.i.d. Normal R.V.s), we find the maximum, and let's call it $M_{1}$. We can repeat this procedure and generate a sequence of maximums: $M_{1}, M_{2}, ..., M_{n}$. This sequence of $n$ values is independent and can be assumed to be identical, considering the underlying generation process is the same. I tried to derive an analytical expression for this and it leads to this. This is basically the distribution of any $M_{i} \ \forall i \in {1, 2, 3..., n}$.

My question is this: what happens to the mean and variance of this distribution as $n \rightarrow \infty$ and the sample size, $k$, is varied?

Law of Large Numbers tells us that these would converge to the Mean and Variance of $M_{i}$ and the underlying distribution is Normal (by CLT). I am stuck at trying to find the mean and variance of $M_{i}$. It's possible to find an analytical expression for the pdf of $M_{i}$, for any general $k$. But, the expressions look difficult to integrate. I tried integrating the pdf to find mean and variance but reached nowhere. A google search revealed that there is a theorem called Extreme Value Theorem EVT, which applies to the case when $k \rightarrow \infty$. Not sure if this specifically applies here.

Also, I tried running some simulations in R and can see that the mean and variance do indeed converge to different values as $k$ is varied. Can someone help in deriving an expression for the mean and variance, or any other insights are much appreciated? I did notice something interesting: as $k$ is increased the mean increases whereas the variance decreases.

$\endgroup$

1 Answer 1

1
$\begingroup$

We can certainly create a table for small $k$, numerically integrating for the mean and variance when the underlying distribution is standard normal; then we can exploit location-scale transformations to get the moments when the underlying distribution is normal with arbitrary mean and variance. Specifically, if $$X_i = \mu + \sigma Z_i \sim \operatorname{Normal}(\mu, \sigma^2)$$ with $Z_i$ standard normal, and $$M_i(k) = \max_{i=1}^k X_i = \mu + \sigma \max_{i=1}^k Z_i = \mu + \sigma M_i^*(k),$$ then computing the mean and variance of $M_i^*(k)$ of the maximum order statistic of the standard normal distribution will suffice. It is not too difficult to compute these to high precision. A table is provided as follows up to $k = 40$: $$\begin{array}{c|cc} k & \operatorname{E}[M_i^*(k)] & \operatorname{Var}[M_i^*(k)] \\ \hline 1 & 0\hphantom{.0000000000000000000} & 1\hphantom{.00000000000000000000} \\ 2 & 0.5641895835477562869 & 0.68169011381620932846 \\ 3 & 0.8462843753216344304 & 0.55946720379736701380 \\ 4 & 1.0293753730039641321 & 0.49171523687474176068 \\ 5 & 1.1629644736405196128 & 0.44753406902066198877 \\ 6 & 1.2672063606114712976 & 0.41592710898324811918 \\ 7 & 1.3521783756069043992 & 0.39191777612675045282 \\ 8 & 1.4236003060452777531 & 0.37289714328672899422 \\ 9 & 1.4850131622092370063 & 0.35735332635781334373 \\ 10 & 1.5387527308351728560 & 0.34434382326069025507 \\ 11 & 1.5864363519080001689 & 0.33324744270295743512 \\ 12 & 1.6292276398719129903 & 0.32363638704764511498 \\ 13 & 1.6679901770491274980 & 0.31520538421231131148 \\ 14 & 1.7033815540999765215 & 0.30773010247051352042 \\ 15 & 1.7359134449410374337 & 0.30104157031389397523 \\ 16 & 1.7659913930547879673 & 0.29500980901031979788 \\ 17 & 1.7939419808826908735 & 0.28953300368769581952 \\ 18 & 1.8200318789687221046 & 0.28453012974137323777 \\ 19 & 1.8444815116038246581 & 0.27993580492832891811 \\ 20 & 1.8674750597983204847 & 0.27569661561853123249 \\ 21 & 1.8891679149213104844 & 0.27176844368099078145 \\ 22 & 1.9096923216814163261 & 0.26811448752380604676 \\ 23 & 1.9291617116425034366 & 0.26470377412772997713 \\ 24 & 1.9476740742256781348 & 0.26151002449149128630 \\ 25 & 1.9653146097535565808 & 0.25851077750621494386 \\ 26 & 1.9821578397613119821 & 0.25568670553246791801 \\ 27 & 1.9982693020065785915 & 0.25302107405446189268 \\ 28 & 2.0137069241232659490 & 0.25049931092298106079 \\ 29 & 2.0285221460475933143 & 0.24810865987769637268 \\ 30 & 2.0427608441715109743 & 0.24583789954688620362 \\ 31 & 2.0564640976381941372 & 0.24367711379799326984 \\ 32 & 2.0696688279289069449 & 0.24161750271345842095 \\ 33 & 2.0824083359701366048 & 0.23965122596881073012 \\ 34 & 2.0947127557684849500 & 0.23777127225118112783 \\ 35 & 2.1066094396039525939 & 0.23597134975445983004 \\ 36 & 2.1181232867564915367 & 0.23424579384730181654 \\ 37 & 2.1292770253732226709 & 0.23258948882088842374 \\ 38 & 2.1400914552352043060 & 0.23099780124849819693 \\ 39 & 2.1505856577287634253 & 0.22946652297472534804 \\ 40 & 2.1607771781750199583 & 0.22799182213242611444 \\ \end{array}$$ Unfortunately, I am not aware of a general closed-form solution for each $k$. We can attempt to fit these, e.g., $$\operatorname{E}[M_i^*(k)] \approx -0.059204467433884 \log ^2 k + 0.79407613941480 \log k + 0.026795590426391, \\ \operatorname{Var}[M_i^*(k)] \approx -0.45226384311138 k^{-2} + 1.23294245728553 k^{-1} + 0.21144333738729,$$ but this is not particularly illuminating. For large $k$, it may be better to do use some other theorems.

$\endgroup$
3
  • $\begingroup$ I guess I will post the original problem that led me to this question. It's from the book Think Stats. It goes like this: there's a baker who claims that his bread loaves are 1000g on average. A statistician suspects foul play and buys loaves from him each day for 1 year and finds that the actual distribution is Normal(950g, 50g). He reports to the police and the baker is warned. The statistician still believes foul play and does the same exercise for another year. Now he finds that the mean is 1000g, but the distribution is skewed. $\endgroup$ Commented May 21, 2020 at 20:16
  • $\begingroup$ Continued: So, the baker tried to manipulate the mean by giving the statistician the heaviest loaf each day. What should be a value of the number of bread loaves the baker must sample each day so that the mean estimated by the statistician at the end of 2nd year is 1000g. I ran some simulations in R and came to an answer of k=4. But, that was literally hit and trial. So, I was wondering if there is a better way to quantify this. The distribution did look like Normal, but I guess it's not that. $\endgroup$ Commented May 21, 2020 at 20:21
  • 1
    $\begingroup$ @SahilGupta Indeed, your value of $k = 4$ is correct, as you can see from the table, since $950 + (50)1.0293 > 1000$ is the smallest value of $k$ for which the mean exceeds $1000$. Unfortunately, as I can see no way to integrate the expectation in closed form, it seems that the only way to get this result is to do it numerically. But perhaps I overlooked some other, more elegant solution. $\endgroup$
    – heropup
    Commented May 22, 2020 at 3:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .