What is the probability that a normal distribution with infinite variance has a value greater than its mean? I got asked something similar to this in interview today.
The interviewer wanted to know what is the probability that an at-the-money option will end up in-the-money when volatility tends to infinity.
I said 0% because the normal distributions that underly the Black-Scholes model and the random walk hypothesis will have infinite variance. And so I figured the probability of all values will be zero.
My interviewer said the right answer is 50% because the normal distribution will still be symmetric and almost uniform. So when you integrate from mean to +infinity you get 50%.
I am still not convinced with his reasoning.
Who is right?
 A: Consider a sequence of normal random variables $X_1, X_2,\ldots, X_n$ with mean $\mu$ and SD $\sigma_n$.
Essentially your interviewer is asking for $\lim_{n\to\infty} P(X_n>\mu)$, given we know $\sigma_n\to \infty$.
Clearly we see $\lim_{n\to\infty} P(X_n>\mu)=\frac{1}{2}$ is independent of $\sigma_n$, which gives us the answer.
Intuitively, instead of conceiving a infinite-variance normal distribution, you should imagine a finite-variance distribution and work with its limits.
A: Neither form of reasoning is mathematically rigorous--there's no such thing as a normal distribution with infinite variance, nor is there a limiting distribution as the variance grows large--so let's be a little careful.
In the Black-Scholes model, the log price of the underlying asset is assumed to be undergoing a random walk.  The problem is equivalent to asking "what is the chance that the asset's (log) value at the expiration date will exceed its current (log) value?"  Letting the volatility increase without limit is equivalent to letting the expiration date increase without limit.  Thus, the answer should be the same as asking "what is the limit, as $t \to \infty$, that the value of a random walk at time $t$ is greater than its value at time $0$?"  By symmetry (exchanging upticks and downticks), (and noting that in the continuous model the chance of being at the money is $0$) those probabilities equal $1/2$ for any $t \gt 0$, whence their limit indeed exists and equals $1/2$.
A: You should be doing your analysis based on a log normal distribution, not a normal one. You interviewer is wrong when he states that the distribution is symmetrical. It would never be, regardless of the variance.  You also need to distinguish between volatility and what you are calling infinite variance. A stocks price, for example,  has no upper limit, thus it has "infinite variance".
