A Barrier-Probability Decomposition for Phoenix Autocall Pricing

# 76948


# Could a phoenix autocall be priced by a snowball option with zero coupon plus expectation of coupons received in knock out observation dates?












I know that coupons in the phoenix autocall can be received in each observation date if the underlying price in that date does not touch down the knock-in barrier and receiving periodic coupons is independent with each other. The phoenix autocall here is simplified to one underlying stock. Also, the coupon barrier is simplified to knock-in barrier. Could I understand the pricing in this way? phoenix option = snowball option with zero coupon rate + sum of expectation of coupons received in all knock-out observation dates = snowball option with zero coupon rate + $\sum_{i=1}^{N_{out}}{c_i e^{-rt_i} P(S_{t_i} > K_{in},\max_{0\leq{u}\leq{t_i}}{S_u}< K_{out})}$, where $N_{out}$ is the number of knock out observation dates, $t_i$ is the $i$th knock out observation date, $K_{in}$ is the knock-in barrier price, and $K_{out}$ is the knock-out barrier price. I split the receiving coupon process and other processes in snowball option pricing and thought the receiving coupon process can be priced by the expectation where the coupons are received in each observation date if the underlying price at that date does not touch down the knock-in barrier and the option does not knock out before or the maximum of underlying price until that date is less than the knock-out barrier. The later probability $P(S_{t_i} > K_{in},\max_{0\leq{u}\leq{t_i}}{S_u}< K_{out})$ can be calculated directly by transferring $S_t$ into $W_t$ and $\max_{0\leq{u}\leq{t}}{S_u}$ into $M_t$ where $M_{t}=\max_{0\leq{s}\leq{t}}{W_{s}}$ is defined as maximum to dates. Below is the procedure. From R. H. Chan et al., Financial Mathematics, Derivatives and Structured Products, P.199, I can obtain the joint pdf of $M_t$ and $W_t$ as $f_{M_{t},W_{t}}(x,y)=\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}1_{x\geq{\max(y,0)}}$. Then the probability to calculate is $$ P(M_{t}<{X}, W_{t}>{Y})$$ $$=\int_{Y}^{\infty}\int_{-\infty}^{X}f_{M_{t},W_{t}}(x,y)dxdy$$$$=\int_{Y}^{\infty}\int_{-\infty}^{X}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}1_{x\geq{\max(y,0)}}dxdy $$. Consider $Y>0$ and $Y\leq0$ two cases. Here, I only consider $X>0$ since the integral is 0 if $X<0$ and obviously $X>Y$.

- $Y>0\Rightarrow{x<X, y > Y>0, x\geq{y}}$

$$ P(M_{t}<{X}, W_{t}>{Y})$$$$=\int_{Y}^{\infty}\int_{-\infty}^{X}f_{M_{t},W_{t}}(x,y)dxdy $$$$=\int_{Y}^{X}\int_{y}^{X}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}dxdy $$$$=\int_{Y}^{X}-\frac{1}{\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}|_{x=y}^{x=X}dy $$$$=\int_{Y}^{X}\left[ -\frac{1}{\sqrt{2\pi{t}}}e^{-\frac{(2X-y)^{2}}{2t}} + \frac{1}{\sqrt{2\pi{t}}}e^{-\frac{y^{2}}{2t}} \right]dy $$$$=-N(X-2X)+N(X)+N(Y-2X)-N(Y)$$$$=2N(X)+N(Y-2X)-N(Y) $$

- $Y\leq0$ $$ P(M_{t}<{X}, W_{t}>{Y})$$$$=\int_{-\infty}^{X}\int_{Y}^{\infty}f_{M_{t},W_{t}}(x,y)dydx $$$$=\int_{-\infty}^{X}\int_{Y}^{0}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}1_{x\geq\max(y,0)}dydx +\int_{-\infty}^{X}\int_{0}^{\infty}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}1_{x\geq\max(y,0)}dydx $$ Calculate $y>0$ and $y\leq0$ separately.

2.1 $Y\leq0,y<0\Rightarrow{x<X,Y<y<0,x\geq0}$

$$ \int_{-\infty}^{X}\int_{Y}^{0}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}1_{x\geq\max(y,0)}dydx$$$$=\int_{Y}^{0}\int_{0}^{X}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}dxdy $$$$ =\int_{Y}^{0}-\frac{1}{\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}|_{x=0}^{x=X}dy $$$$=\int_{Y}^{0}\left[ -\frac{1}{\sqrt{2\pi{t}}}e^{-\frac{(2X-y)^{2}}{2t}} + \frac{1}{\sqrt{2\pi{t}}}e^{-\frac{y^{2}}{2t}} \right]dy$$$$ =-N(-2X)+N(0)+N(Y-2X)-N(Y) $$

2.2 $Y\leq0,y\geq0\Rightarrow{x<X,y\geq0,x\geq{y}}$ $$ \int_{-\infty}^{X}\int_{0}^{\infty}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}1_{x\geq\max(y,0)}dydx$$$$=\int_{0}^{X}\int_{y}^{X}\frac{2(2x-y)}{t\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}dxdy $$$$=\int_{0}^{X}-\frac{1}{\sqrt{2\pi{t}}}e^{-\frac{(2x-y)^{2}}{2t}}|_{x=y}^{x=X}dy $$$$=\int_{0}^{X}\left[ -\frac{1}{\sqrt{2\pi{t}}}e^{-\frac{(2X-y)^{2}}{2t}} + \frac{1}{\sqrt{2\pi{t}}}e^{-\frac{y^{2}}{2t}} \right]dy $$$$=-N(X-2X)+N(X)+N(-2X)-N(0) $$ Adding these two together, we have $$ P(M_{t}<{X}, W_{t}>{Y})$$$$=-N(-2X)+N(0)+N(Y-2X)-N(Y)-N(X-2X)+N(X)+N(-2X)-N(0)$$$$=2N(X)+N(Y-2X)-N(Y) $$ for $Y\leq0$ case. Finally, both $Y>0$ and $Y\leq0$ give the same result $$P(M_{t}<{X}, W_{t}>{Y})=2N(X)+N(Y-2X)-N(Y)$$. Thus, we can calculate the probability and then the expectation. My questions are

- Is this way of pricing a phoenix option reasonable? Or could it be used in reality? All thing begins with solving PDE by finite difference method since I priced the snowball option by splitting it into four options (Autocall + Up-Out and Down-Out - Up-Out Put + Up-Out and Down-out Put) and now I began to consider pricing the phoenix options.

- If this way of pricing is reasonable, is my procedure of calculating the joint probability correct? I am not sure.

- If this way of pricing is not correct, how should I price the phoenix option with one underlying except Monte Carlo simulation? Greatly appreciate your helps!!!