# Given the density function of $S^{1}$ in one-period model, find the risk-neutral measure

Consider the one period market model $$\left(\overline{\pi},\overline{S}\right)$$ consisting of a risk-free asset $$\left(\pi^{0},S^{0}\right)=(1,1+r)$$ and a risky $$\left(\pi^{1},S^{1}\right)$$

Let $$r > -1$$ and $$\pi^{1}>0$$. $$S^{1}$$ has strictly positive density function $$f: (0,\infty)\to \left(0,\infty\right)$$, i.e. $$P(S^{1}\leq x)=\int_{0}^{x}f(y)\mathrm{d}y$$ for $$x > 0$$.

Find the risk-neutral measure $$\mathbb Q$$.

My idea:

I think we need to characterize $$\mathbb Q$$ in terms of its Radon-Nikodym derivative with respect to $$\mathbb P$$, i.e. $$\frac{\mathrm{d}\mathbb Q}{\mathrm{d}\mathbb P}$$. By definition of being risk-neutral, $$\mathbb Q$$ has to satisfy:

$$(1+r)\pi^{1}=E_{\mathbb Q}[S^{1}]$$

By the Radon-Nikodym derivative approach, we get

$$(1+r)\pi^{1}=E_{\mathbb Q}[S^{1}]=E_{\mathbb P}\left[S^{1}\frac{d\mathbb Q}{d\mathbb P}\right]=\int_{\Omega}S^{1}(\omega)\frac{\mathrm{d}\mathbb Q}{\mathrm{d}\mathbb P}(\omega)\mathbb P(\mathrm{d}\omega)=\int _{0}^{\infty}\left(S^{1}\circ f\right)(y)\left(\frac{d\mathbb Q}{d\mathbb P}\circ f\right)(y)f(y)\mathrm{d}y$$

I have no idea how to proceed to get an expression of $$\frac{\mathrm{d}\mathbb Q}{\mathrm{d}\mathbb P}$$. Any ideas? Am I on the right track?

• I don’t see how this can be achieved. There’s not enough information to define a unique risk neutral measure I think
– dm63
Commented Dec 18, 2020 at 15:35
• I added the source of the exercise. Commented Dec 18, 2020 at 15:45