# Value of contingent claim at a given time

Consider a contingent claim whose value at maturity T is given by

$$\min(S_{T_0}, S_T)$$

where $$T_0$$ is some intermediate time before maturity, $$T_0 < T$$, and $$S_T$$ and $$S_{T_0}$$ are the asset price at $$T$$ and $$T_0$$, respectively. Assuming the usual Geometric Brownian process for the price of the underlying asset that pays no dividend, show that the value of the contingent claim at time $$t$$ is given by

$$C(T-T_0, s) = S[1-N(d_1)+e^{r(T-T_0)}N(d_2)]$$

where S is the asset price at time $$t$$ and

$$d_1 = \frac{\left( r+ \frac{\sigma^{2}}{2} \right)(T-T_0)}{\sigma \sqrt{T-T_0}}, d_2 = d_1 - \sigma \sqrt{T-T_0}$$

I asked a similar question like this before but I am confused now when the min function is applied. Any suggestions is greatly appreciated.

This is related to the valuation of a forward start option. Let's assume Black-Scholes world (constant vol) and for simplicity byt without loss of generality I'll set $$r=q=0$$.
Notice that $$\min(S_T,S_t) = S_T - (S_T - S_t)_+$$ The value of this claim at time $$0$$ is \begin{align} E_0 \left[ \min(S_T,S_t) \right] &= E_0\left[S_T \right] - E_0 \left[(S_T - S_t)_+\right] \\ &= S_0 - E_0 \left[(S_T - S_t)_+\right] \end{align}
The second term is a Type II ATM forward start option. To value it use the 'conditioning trick': \begin{align} E_0 [\left[(S_T - S_t)_+\right] &= E_0 \left[ E_t \left[(S_T - S_t)_+\right] \right] \\ &= E_0 [C(S_t,S_t,T-t)] \\ &= E_0 [S_t C(1,1,T-t)] \\ &= S_0 C(1,1,T-t) \end{align} with $$C(1,1,T-t)$$ the Black-Scholes call price function with spot price $$1$$ and strike $$1$$.
So, $$E_0 \left[ \min(S_T,S_t) \right] = S_0 \left(1 - C(1,1,T-t) \right)$$