Is it true that every tradable asset must have a log-normal dynamics under the risk neutral measure where the drift term is the short rate $r$? I.e., is it true that if $X$ is a tradable asset then $$\frac{\mathrm{d}X(t)}{X(t)} = r(t)\mathrm{d}t + \sigma(t, X(t))\mathrm{d}W(t),$$ where $W$ is a Brownian motion under the risk neutral measure for some $\sigma$ (which may or may not be deterministic)?
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1$\begingroup$ If $X$ is a tradable and you want to price a derivative on $X$, then yes you need $r$ as the drift, i.e. you need to price the claim as if the drift of $X$ is $r$. $\endgroup$ – ilovevolatility Oct 7 '20 at 16:47
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$\begingroup$ My question is not about the drift term, which has to $r$, but about the log-normality of the process. In principle it could be something of the form $dX(t) = r(t)dt + \sigma(t, X(t))dW(t)$ right? $\endgroup$ – desmodena Oct 7 '20 at 17:39
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1$\begingroup$ As you clearly know the drift term must be $r$ then you also know that the expected return of a tradable under the risk neutral measure should be the same as the expected return of the money market account, right? There is a reason that the drift term must be $r$. $\endgroup$ – ilovevolatility Oct 7 '20 at 17:54
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2$\begingroup$ My point being the $r$ is there because if $X$ is a tradable it must be the case that $dX = rXdt + \sigma(X,t) dW$ so that the expected return of the asset under RN is the same as the money market account to prevent arbitrage. That is the reason for the $r$. So no, it cannot take just any form, but it doesn't have to be lognormal either. $\endgroup$ – ilovevolatility Oct 7 '20 at 18:04
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$\begingroup$ I'd say that as long as the discounted price process is a martingale, it can be any distribution, doesn't have to be log-normal. $\endgroup$ – Jan Stuller Oct 7 '20 at 18:41
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Our market has a tradeable asset $S$ and a risk-less money market account $B$, that is, the numéraire of the risk-neutral measure. We assume the following standard conditions, which are widely applicable to most common models:
- We work in an Itô diffusion setting, and neglect jump modelling: $$\begin{align} & dS_t=\mu(t,S_t)dt+\sigma(t,S_t)dW^S_t \\ & dB_t=r(t,B_t)dt + \varsigma(t,B_t)dW^B_t \end{align}$$
- The money market account $B$ has no stochastic contribution (because it is riskless): $$\varsigma \equiv0$$
- Local martingales are martingales.
The exact requirement is that the discounted asset price is a martingale under the risk-neutral measure, and we want to determine the expression of the drift term $\mu(\cdot)$ $-$ note that all dynamics are expressed under the risk-neutral measure. Per our assumptions, this is equivalent to saying that there exists a function $\eta(\cdot)$ and a Brownian Motion $W$ such that: $$d\left(\frac{S_t}{B_t}\right)=\eta(t,S_t,B_t)dW_t$$ Applying Itô's Lemma: $$\begin{align} d\left(\frac{S_t}{B_t}\right) &=\frac{1}{B_t}dS_t-\frac{S_t}{B_t^2}dB_t+\frac{S_t}{B_t^3}d[B,B]_t-\frac{1}{B_t^2}d[S,B]_t \\ &=\frac{1}{B_t}dS_t-\frac{S_t}{B_t^2}dB_t \end{align}$$ In order to cancel the drift contributions in the above equation, we need to have: $$\begin{align} \mu(t,S_t)=r(t,B_t)\frac{S_t}{B_t} \end{align}$$ That is, the drift of the asset needs to be equal to the drift of the money market account, adjusted by the price ratio between the asset and the MMA:
- For example, if $B$ is exponential, that is $r(t,B_t)=rB_t$, then: $$\mu(t,S_t)=rS_t$$
- However, if $B$ is linear, that is $r(t,B_t)=r$, then: $$\mu(t,S_t)=r\frac{S_t}{B_t}$$ In this case, note that $B_t=B_0+rt$, so if $B_0=0$, we have the following, rather peculiar drift: $$\mu(t,S_t)=\frac{S_t}{t}$$
In practice, the money market account is always assumed to have an exponential form because it is the most sensible way to represent mathematically such a security. Because the dynamics of the money market account will restrict the drift of the asset (if we are to ensure the martingale requirement), the drift of the asset will be $rS_t$ in most models. However, the model might not necessarily be log-normal. For example, the Bachelier model is usually specified as follows: $$dS_t=rS_tdt+\sigma dW_t^S$$ which corresponds to a normal distribution for $S$.
answered Oct 7 '20 at 17:48
