Why the Black-Derman-Toy Scale Parameter Is the Short-Rate Median

Summary

The exchange explains why the scale parameter U(t) in a stated Black-Derman-Toy short-rate representation is the median of the short-rate distribution at time t. The rate is expressed as U(t) multiplied by an exponential term involving volatility and Brownian motion. Since Brownian motion at a fixed time is symmetrically distributed around zero, the probability that it is nonnegative is one half. For positive volatility, the exponential factor is at least one exactly when the Brownian term is nonnegative, so the rate is at least U(t) with probability one half.

This gives the median interpretation under the representation and distributional assumptions in the question. The answer is a concise probability argument, not a general derivation of the Black-Derman-Toy model or its calibration procedure. The result also relies on the usual continuous Brownian distribution, so equality at the threshold has probability zero; it does not establish that U(t) is the mean or another measure of central tendency.

Key ideas

• The short rate is represented as a scale parameter times an exponential Brownian factor.
• Brownian motion at a fixed time is symmetrically distributed around zero.
• With positive volatility, the rate is at least U(t) when the Brownian term is nonnegative.
• That event has probability one half, supporting U(t) as the median under the stated model.

Tags

# Jamshidian's formulation of Black Derman Toy


# Jamshidian's formulation of Black Derman Toy












In his 1991 paper on forward induction of binomial tree calibration on BDT model, it is stated that

$$r(t) = U(t)\exp(\sigma(t)W(t))$$

where $r$ is the short rate modelled by Black Derman Toy. It is further stated that $U$ above is the median of short rate distribution at time $t$. Now if I recall correctly, in BDT model the short rate is modelled as $$d\log(r(t)) = \left[\theta(t) - \frac{\sigma'(t)}{\sigma(t)}\log(r(t))\right]dt + \sigma(t)dW$$ Now I can see that $r(t)$ might be put into the form above, but how is the claim that $U(t)$ being the median is justified?

After some google searches, I can't seem to find a reference for this.

## Answer by NN2 (score 1, accepted)

https://quant.stackexchange.com/a/79700

It suffices to prove $\mathbb{P}(r(t) \ge U(t))=\frac{1}{2}$ which is true because $$\begin{align} \mathbb{P}(r(t) \ge U(t))&=\mathbb{P}(\exp{\sigma(t)W(t)} \ge 1) = \mathbb{P}(W(t) \ge 0) = \frac{1}{2} \end{align}$$

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