Discounted Asset Prices as Martingales Under the Risk-Neutral Measure

Summary

The document asks how to derive the price dynamics of a claim whose payoff depends on the short rate, and how to show that its price divided by the money-market account is a martingale under the risk-neutral measure. The answer applies Itô’s rule to the discounted price. If the claim price has a drift equal to the short rate times its price, that drift cancels the growth of the bank account, leaving only a stochastic-integral term.

This gives the standard intuition for risk-neutral valuation: discounted asset prices have no drift under the pricing measure, and suitable integrability conditions allow the stochastic integral to be a martingale. However, the response assumes the price dynamics rather than deriving them from the payoff and short-rate model. Its notation switches between Brownian motions and does not state the regularity or integrability conditions needed for a rigorous martingale result.

Key ideas

• The money-market account grows at the prevailing short rate.
• Discounting a price process with matching short-rate drift cancels its drift term.
• The resulting discounted price is a stochastic integral under the risk-neutral measure.
• A true martingale conclusion requires suitable integrability conditions.
• The provided derivation assumes the claim price dynamics instead of deriving them from the payoff.

Tags

# Normalized price process $Z(t)=\frac{\Pi(t)}{B(t)}$


# Normalized price process $Z(t)=\frac{\Pi(t)}{B(t)}$












If an interest rate model with the following $P$-dynamics for the short rate.

$$dr(t)=\mu(t,r(t))dt+\sigma(t,r(t))d\bar{W}(t)$$

Now consider a $T$-claim of the form $\chi = \Phi(r(T))$ with corresponding price process $Π(t)$.

Can anyone help me to find stochastic differential of $Π(t)$ ?

and show that the normalized price process

$$Z(t)=\frac{\Pi(t)}{B(t)}$$

is a $Q$-martingale.?

I appreciate any help.

Thanks.

## Answer by user16651 (score 1, accepted)

https://quant.stackexchange.com/a/18571

You Know that $dB_t=r_tB(t)dt$ . Ito's formula give us \begin{align} dZ(t)=\frac{1}{B(t)}d\,\Pi(t)-\frac{\Pi(t)}{B\,^2(t)}dB(t)+0 \end{align} As your teacher mentioned, $d\Pi(t)=r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t)$,Thus we have \begin{align} & dZ(t)=\frac{1}{B(t)}[r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t)]-\frac{\Pi(t)}{B\,^2(t)}r(t)B(t)dt\\ & dZ(t)=\frac{1}{B(t)}r(t)\Pi(t)dt+\frac{1}{B(t)}\sigma(\Pi(t),t)dW(t)-\frac{1}{B(t)}r(t)\Pi(t)dt\\ \end{align} then \begin{align} dZ(t)=\frac{1}{B(t)}\sigma(\Pi(t),t)dW(t) \end{align} Martingale Representation Theorem shows that $Z(t)$ is a Martingale.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.