# 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?

• Actually there is a (weak) limit of normal distributions as variance increases to infinity. It involves a forbidden infinitesimal 1/Aleph(0). You can read my article about infinitesimals in Research Gate or in Academia. Type "H. Tomasz Grzybowski" in Google, get to the Research Gate page with my articles, click "Contributions" and find it. Aug 24 '18 at 12:23
• Welcome to our site, @H.TomaszGrzybowski. I have converted your post to a comment because I knew you hadn't yet accrued the reputation to create a comment, but it doesn't actually answer the question and therefore cannot remain as a reply. It would be interesting to read a solution to this problem that is based on your idea of infinitesimals and a weak limit. Do you still arrive at the value of $1/2$ or do you find the value is undefined?
– whuber
Aug 24 '18 at 12:46

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 $1/2$.

• +1 In short, physical reasoning: two possible outcomes, perfectly symmetric, and probabilities of all possible outcomes must sum up to 1 -- the only answer can be 1/2 (-;
– user88
Mar 14 '11 at 19:48

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.

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".

• It is correct that a lognormal distribution is involved, but it is unnecessary to invoke it, as my reply shows. The underlying normal distribution is symmetrical of course. The fact that a stock price (or anything else) has no upper limit does not imply its distribution has infinite variance. In the Black-Scholes theory, by the way, volatility is indeed the variance parameter (for the distribution of logarithms).
– whuber
Mar 14 '11 at 18:32
• we consider the option, not the stock.
– Wok
Mar 14 '11 at 18:56
• @wok True, but the theory depends on the distribution of the asset (stock) prices. The distribution of the option values is neither normal nor lognormal.
– whuber
Mar 14 '11 at 19:51