I'm working through the following little exotic exercise and have some questions and curiosity as to whether I'm on the right track
Consider the claims $$Y_t=\frac{1}{S_t}$$ $$X=\frac{1}{S_T}$$ a) Can $Y_t$ be the arbitrage-free price of a traded derivative?
Answer?-- So this question is for some reason stumping me. I suppose it means the literal process $Y_t$ (that is, not under a risk-neutral expectation), which seems highly unlikely to be an arb free price process. I just can't seem to put it in any rigorous terms.
b) Derive an expression for the arbitrage free price process $\pi_t[X]$
Under risk-neutral valuation, we have $$\pi_t[X]=E^Q[\frac{X}{B_T}]=E^Q[\frac{\frac{1}{S_T}}{B_T}]=E^Q[\frac{1}{S_TB_T}]$$ So, here's where I had the idea to multiply both sides by $S_t$. Now, I've done a lot of problems with change of numeraire, but this really isn't that, so I'm now going to continue under the assumption that we are still under Q: $$\pi_t[X]=\frac{1}{S_t}E^Q[\frac{S_t}{S_TB_T}]<=>$$ $$\pi_t[X]=\frac{1}{S_t}E^Q[e^{-r(T-t)+(\frac{1}{2}\sigma^2-r)(T-t)-\sigma(W_T-W_t)}]<=>$$ $$\pi_t[X]=\frac{1}{S_t}E^Q[e^{(\frac{1}{2}\sigma^2-2r)(T-t)-\sigma(W_T-W_t)}]$$ Using the fact that $E[e^{\mu+\sigma Z}]=e^{\mu+\frac{1}{2}\sigma^2}$, we have $$\pi_t[X]=\frac{1}{S_t}e^{(\sigma^2-2r)(T-t)}$$