the distribution having p.d.f.
f(x)=exp(a-x) x>0
Let x1=min{x1,x2,...,xn}and
zn=n(x1-a), find the limiting distribution of zn
To find the limiting distribution of zn, we need to find the cumulative distribution function (CDF) of zn and then take the limit as n approaches infinity.
First, let's find the CDF of zn:
CDF(zn) = P(zn ≤ z)
= P(n(x1 - a) ≤ z)
= P(x1 ≤ z/n + a)
Since we know that x1 is the minimum of n random variables, we can express P(x1 ≤ z/n + a) in terms of the individual random variables xi's.
P(x1 ≤ z/n + a) = 1 - P(x1 > z/n + a)
= 1 - P(x1 > z/n + a, x2 > z/n + a, ..., xn > z/n + a)
Since the random variables xi's are independent, we can express this probability as a product of individual probabilities:
P(x1 ≤ z/n + a) = 1 - P(x1 > z/n + a) * P(x2 > z/n + a) * ... * P(xn > z/n + a)
The random variable x has a probability density function (PDF) of f(x) = exp(a - x) for x > 0. Therefore, the probability of a random variable xi being larger than z/n + a can be expressed as:
P(xi > z/n + a) = integral of f(x) from z/n + a to infinity
= integral of exp(a - x) from z/n + a to infinity
= exp(a - (z/n + a))
= exp(-z/n)
Substituting this into the expression for P(x1 ≤ z/n + a), we get:
P(x1 ≤ z/n + a) = 1 - exp(-z/n)^n
Now let's take the limit as n approaches infinity:
lim(n→∞) (1 - exp(-z/n)^n)
Since exp(-z/n) approaches 1 as n approaches infinity, the limit becomes:
lim(n→∞) (1 - 1^n)
Which simplifies to:
1 - 1
Therefore, the limiting distribution of zn is:
CDF(zn) = 0
In other words, the limiting distribution of zn is a degenerate distribution, where the probability is concentrated at a single point, which is 0.