The question is as follows, suppose (${z_0,z_1,z_2..z_n}$) are independent random variables drawn from the same distribution F.Define the minimum of this set has a ($a=\min(z_0,z_1..z_n)$).I want to show that the probability of next drawn from the distribution (so that represents $z_{n+1}$) being inferior to $a$ is inferior to $\frac{1}{n}$
Show that: $$Prob(z<a) < \frac{1}{n} $$
Things that I have tried and dueled, it is not difficult to see that the minimum has the pdf ($Prob(A<a) =1-(1-F(a)))^n$ so it has a probability density function off $nf(a))(1-F(a))^{n-1}$
So to explicitly compute the required probability we can do something as follows: and calculate : $$ Prob(z<a) =\int_{z=-\infty}^{z=+\infty}\int_{a=z}^{a=+\infty} f(a)f(z)dadz=\int_{z=-\infty}^{z=+\infty}\int_{a=z}^{a=+\infty} nf(z)f(a)(1-F(a))^{n-1}dadz$$
the only bound that I can take from this is putting F(a) = 0 in the integral witch gives
$$ Prob(z<a) <= n\int_{z=-\infty}^{z=+\infty}\int_{a=z}^{a=+\infty} f(z)f(a)dadz = n\int_{z=-\infty}^{z=+\infty}f(z)[F(+\infty)-F(z)]dz= n\int_{z=-\infty}^{z=+\infty}f(z)[1-F(z)]=n(1-\int_{z=-\infty}^{z=+\infty}f(z)F(z)dz) =\frac{n}{2}$$
evaluating this integral by parts: $$\int f(z)F(z)dz= F(z)^2-\int f(z)F(z)dz$$ which gives $$\int f(z)F(z)dz = \frac{F(z)^2}{2} $$ evaluating it in extremes that gives $\frac{1}{2}$ . So in my defense the value for n=2 seems nice xp
I believe I am attacking the problem in the wrong way , any suggestion would be very helpful. (note I also am not sure if I formulated correctly the problem).
Well I guess that now I solve it , if i don't say (F(a) = 0),Its visible that that integrals can be simplified to arrive to the right answer. you can integrate $f(a)(1-F(a))^n$ rather easily , and then you have another similar integral and you arrive at $\frac{1}{n+1}$ still probably was a overkil
$P(z<a)$ is the same as the probability that $z_0,\dots,z_n,z_{n+1}$ has a unique minimum value $z_{n+1}.$ But if the values $z_0,\dots,z_n,z_{n+1}$ come from identical independent random variables, then given that there is a unique minimum, the probability that it is $z_{n+1}$ is $\frac{1}{n+2}.$
So if $q$ is the probability that $n+2$ of these i.i.d. have a unique minimum, then:
$$P(z_{n+1}<a)=\frac{q}{n+2}\leq \frac{1}{n+2}\leq\frac{1}{n}$$
Note that when the distribution is continuous, then the probability that two of the $z_i$ are equal is zero, and thus we have $q=1$ and $P(z_{n+1}<a)=\frac{1}{n+2}.$