# Finding the limit $\lim_{n \to \infty} P_0^n$ for a European Cash-or-Nothing put option with $P=K^2\cdot \mathbf{1}_{\{S_T < K\}}$

Exercise :

Let $$K>0$$. A European Cash-or-Nothing put option $$P$$ has the following pay-out profile : $$P=K^2\cdot \mathbf{1}_{\{S_T < K\}}$$ Let $$P_0^n$$ be the no-arbitrage value at time $$0$$ and the $$n$$-th binomial model. Calculate the limit $$\lim_{n \to \infty} P_0^n$$. You can use fact that the random variable $$S_T^n$$ follows the log-normal distribution when $$n \to \infty$$ and that $$\lim_{n \to \infty} \mathbb{E}_\mathbb{Q}[\mathbf{1}_{\{S_T < K\}}] = \Phi(-d_2)$$.

Note : For the binomial models, it is : $$S_T^n = \prod_{i=1}^n(1+R_i^n)$$

Question :

Now, I know how one shows what the limit of $$P_0^n$$ is when simply $$P=K$$, but I am really struggling to manipulate the case of this special given $$P$$. Any help will be greatly appreciated.

• I think a good starting point would be to use the Monotone Convergence Theorem en.wikipedia.org/wiki/Monotone_convergence_theorem – FunnyBuzer Feb 19 '19 at 12:54
• @FunnyBuzer Hi. I don't know how to use the $P$ given. It's not standard as when we say it pays off $K$. Can you help me with the initial expression of $P_0^n$ ? – Rebellos Feb 19 '19 at 12:58
• Not sure I understand you comment. You do know how to solve it when the payoff is $K \mathrm{1} \left\{ S_T < K \right\}$ and struggle with it being $K^2$? – LocalVolatility Feb 19 '19 at 17:09
• @LocalVolatility Yes. We proved the limit of $P_0^n$ at class when the payoff is simply $K$ by using the expected martingale value of $(K-S_T)^+$ and expanding using the Triangular Central Limit Theorem, followed by the definition integrals and leading to the expressions involving $d_1$ and $d_2$. But, how would $K^2$ change that ? Would it just be the same but with $(K^2-S_T)^+$ ? – Rebellos Feb 19 '19 at 17:11
• I think you are overcomplicating things. An option that pays $K^2 \mathrm{1} \left\{ S_T < K \right\}$ has expected payoff which is just $K$ times that of an option that pays $K \mathrm{1} \left\{ S_T < K \right\}$. It is just a constant that you can take out of the expectation. Also, I don't see why your starting point is the call payoff as opposed to just the payoff the product you are interested in. – LocalVolatility Feb 19 '19 at 17:14