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## Saturday, January 4, 2014

### Pr(S_T > K | S_0 > K and r=0), intuitively

The original question -- "Assuming S_0 > K and r = 0, denote C := time-0 value of a binary call. What happens to C as ttl -> 0 or ttl -> infinity. Is it below or above 0.5?"

C = Pr(S_T > K), since the discounting to PV is non-issue. So let's check out this probability. Key is the GBM and the LN bell curve.

We know the bell curve gets more squashed  to 0 as ttl -> infinity. However, E S_T == S_0 at all times, i.e. average distance to 0 among the diffusing particles is always equal to S_0. See http://bigblog.tanbin.com/2013/12/gbm-with-zero-drift.html

 together with the median. Eventually, the median will be pushed below K. Concrete illustration -- S_0 = \$10 and K = \$4. As TTL -> inf, the median of the LN bell curve will gradually drop until it is below K. When that happens, Pr (S_T > K) < 50%. In fact, C -> 0 as ttl -> infinity.

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ttl -> 0. The particles have no time to diffuse. LN bell curve is narrow and tall, so median and mean are very close and merge into one point when ttl -> 0. That means median = mean = S_0.

By definition of the median, Pr(S_T > median) := 0.5 so Pr(S_T > S_0) = 0.5 but K is below S_0, so Pr(S_T > K) is high. When the LN bell curve is a thin tower, Pr(S_T > K) -> 100%