1
$\begingroup$

I have to prove something about normaly distributed variables.

Let $X_1, X_2, ... $ be i.i.d. with normal distribution $N(\mu, \sigma^2)$, where $\mu >0$. Define: $S_n := X_1 + X_2 + ... + X_n$ and $Z_n := \max\{S_0, S_1, S_2, ..., S_n\}$, where $S_0=0$.

I must show that random variable $Z_{\infty} = \max\{S_0, S_1, S_2, ...\}$ is almost surely finite, which means that $P(Z_{\infty} < \infty)=1$.

Any help will be very appreciated. Thanks.

$\endgroup$
  • $\begingroup$ @Sam $Z_{\infty}<1$....? That's far from being true... $\endgroup$ – saz Feb 12 '17 at 18:46
  • $\begingroup$ Continuous probability distributions are normalised to one.$ \int^{\infty}_{0} f(x) dx = 1 $ This means that $ P(Z_{\infty} < \infty) < 1 $ however it's expectation value will have the same weight, due to some function g(x) that generates the values in $ N(\mu ,\sigma^{2}) $ So $ g(x) $ may simply be a value of x.$ \int_{a}^{b}g(x_{n})f(x)dx = X_{n} $ $\endgroup$ – McTaffy Feb 12 '17 at 18:51
  • $\begingroup$ @saz corrected. $\endgroup$ – McTaffy Feb 12 '17 at 18:52
0
$\begingroup$

This is wrong. In fact, we know by the strong law of large numbers that $S_n/n \to \mu$ as $n \to \infty$. Under your assumption that $\mu>0$ it follows that $S_n \sim n\mu$ so that $S_n \to + \infty$ and thus $P(Z_{\infty}=\infty)=1$.

My guess is that the problem is stated incorrectly, because the statement becomes true if you reverse signs. In other words, it is true that $P(\inf_n S_n>-\infty) = 1$. The reason for this is exactly the same: because $S_n/n \to \mu>0$ as $n \to \infty$, so that $S_n$ can be negative for only finitely many $n$ (almost surely).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.