0

If $F(x) = 1 - \frac{1}{2}x^2$ for $x \in [-1, 0)$ is the CDF of independent $X_i$, then what does $\max_{1 \leq i \leq n} X_i$ converge to as $n \rightarrow \infty$? ($F(x) = 0$ for $x < -1$ and $F(x) = 1$ for $x \geq 0$.)

I don't recognise the continuous distribution.

If the maximum is at most x, then all $n$ $X_i$ are at most $x$, which gives me $(1 - \frac{1}{2}x^2)^n$. How do you work out the limit as $n \rightarrow \infty$?

johnsmith
  • 343

1 Answers1

1

For $x \in [-1,0)$, $1-\frac 1 2x^{2}$ lies between $\frac 1 2 $ and $1$. Note that $t^{n} \to 0$ as $n \to \infty$ if $0\le t<1$. So the limiting CDF is given by $F(x)=0$ for $x<0$, $1$ for $x \ge 0$. This corresponds to the constant random variable $X=0$.

geetha290krm
  • 36,632