Central limit theorem maximum

Question

$ξ_1, ξ_2, ξ_3.... ξ_n.... $ they are equally distributed, independent, and have a single expectation and a single variance. Let $M_n = max(ξ_1, ξ_1 + ξ_2,...,ξ_1 + ξ_2 + ... + ξ_n) $ .

Is it possible to find for every ${x}$ where $\lim \limits_{n\to\infty}\mathbb{P}(M_n \leq n + x\sqrt n)$ ?
There was a tip: Central limit theorem would help
Seems that it is simple trick, but I have no idea =( I would be very grateful for help!

Answer 1

Have a look at the Darling Erdos Theorems. They are reviewed for example in this paper:

https://projecteuclid.org/journals/annals-of-probability/volume-26/issue-2/Darling-Erd%c5%91s-theorems-for-normalized-sums-of-iid--variables/10.1214/aop/1022855652.full

Answer 2

The running maximum formula in Wikipedia only works if there is no drift. If the mean is not 0, use equation (1.5) from Shepp, Lawrence A. "The joint density of the maximum and its location for a Wiener process with drift." Journal of Applied probability (1979): 423-427. First integrate over $x$ to get the bivariate density. There doesn't seem to be a closed form for the marginal density of $Y$ unless the drift parameter ($c$) is $0$.

This R program will evaluate the distribution function by numerical integration. I include an example where the random variables are iid normal with mean 0.5 and standard deviation 2. Then, the graph shows the empirical distribution function (black) from the simulated values and the distribution evaluated by numerical integration (red).

library(pracma)
library(cubature)

fYth=function(x,c1,sig) {
  y=x[1]
  th=x[2]
  res= (
    2*exp((c1^2*(1 - th + 2*th^2))/(2*sig^2*(-1 + th)) + (y*((2*c1)/(-1 + th) + y/th))/(2*sig^2) - 
            (c1^2*th^3 - c1*(-2 + th)*th*y + (-1 + th)*y^2)/(sig^2*(-1 + th)*th))*sig*y-
      (c1*exp((c1^2*th*(1 + th))/(2*sig^2*(-1 + th)) + (y*((2*c1)/(-1 + th) + y/th))/(2*sig^2) - (c1^2*th^3 - c1*(-2 + th)*th*y + (-1 + th)*y^2)/
                (sig^2*(-1 + th)*th))*sqrt(2*pi)*sqrt(1 - th)*y)+
      c1*exp(-((c1^2*th^3 - c1*(-2 + th)*th*y + (-1 + th)*y^2)/(sig^2*(-1 + th)*th)) + ((2*c1*th^2*(c1 + y))/(-1 + th) + (-(c1*th) + y)^2)/
               (2*sig^2*th))*sqrt(2*pi)*sqrt(1 - th)*y*erf((c1*sqrt(1 - th))/(sqrt(2)*sig))
  )/(2*pi*sig^3*th*sqrt((1 - th)*th))
  return(res)
}

FY=function(y,c1,sig) {
  a1=adaptIntegrate(fYth,lowerLimit=c(0,0.000001),upperLimit =c(y,0.999999), 
                    c1=c1,sig=sig,maxEval=10000)
  if (is.na(a1$integral) | a1$integral>1) print(c(y,a1$returnCode,a1$integral))
  return(a1$integral)
}

n=1000
nsim=10000
mu=0.5
sig=2
x=matrix(rnorm(n*nsim,mu,sig),ncol=n)
y=apply(x,1,function(x1) max(cumsum(x1)))
plot(ecdf(y))
pts=FYs=c(350:700)
for (i in 1:length(pts)) FYs[i]=FY(pts[i],mu*n,sig*sqrt(n))
lines(pts[!is.na(FYs)],FYs[!is.na(FYs)],col="red")