I want to prove this statement that looks easy, but I got stuck. Please kindly help me.
Suppose that we are in the Black-Scholes framework and that $r=0$ so that the stock price dynamic, under the risk neutral measure $Q$, is given by $$dS_t=\sigma S_t dW_t$$
Suppose the time now is $0$. Consider the option whose payoff at time $T$ is $$V_T=\left(S_T-\min_{0 \le t \le T }S_t\right)^+$$
Let $R$ be defined by the likelihood ratio $$\frac{dR}{dQ} = \frac{S_t}{S_0}$$ on $F_t$ , $0 \le t \le T$.
Show that $$\frac{1}{S_0} E^Q[V_T] = E^R\left[\left(1- \min_{0 \le t \le T} P_t\right)^+\right]$$ where $P_t = \frac{S_0}{S_t}$ for all $t \ge 0 $.
I tried for a long time but only managed to get the following:
Using Radon Nikodym derivative process,
$$E^R[V_T] = E^Q\left[\frac{S_T}{S_0} V_T\right]$$
which implies $$\frac{1}{S_0} E^Q[V_T] = \frac{1}{S_T}E^R[V_T]$$ $$=E^R\left[1-\frac{\min_{0 \le t \le T} S_t}{S_T}\right]$$
I know I am supposed to use the fact: $$\min_{0\le t \le T} f(T-t) = \min_{0\le T-u\le T} f(u) = \min_{0\le u\le T}f(u)$$ (with $u =T-t$).
But I have no idea how I can derive that very last part. Thanks in advance!