Continuous Limit for a Cash-or-Nothing Put in a Binomial Model

Summary

The exercise asks for the limiting no-arbitrage value of a European cash-or-nothing put as the number of binomial steps grows. Its payoff is specified as K squared when the terminal asset price is below the strike and zero otherwise. The prompt supplies the limiting risk-neutral probability of finishing below the strike, expressed through the standard Black–Scholes quantity Φ(−d₂). Thus, the limiting valuation follows by multiplying the discounted payoff amount by that probability, assuming the usual risk-neutral pricing setup and discounting convention.

The answer states a result proportional to the normal cumulative probability, but gives K rather than K squared and does not show discounting. That appears inconsistent with the payoff written in the exercise unless an unstated normalization or convention is intended. The document gives no derivation of the binomial convergence or parameter definitions, so the stated expression should be checked against the specified payoff and model assumptions before use.

Key ideas

• As the binomial step count increases, the terminal asset price converges to a lognormal distribution under the stated setup.
• The payoff event is that the terminal price finishes below the strike.
• The limiting risk-neutral event probability is expressed as Φ(−d₂).
• Pricing requires multiplying the payoff amount by the event probability and applying the relevant discounting.
• The provided answer uses K despite the exercise specifying a K-squared payoff, so its scaling requires verification.

Tags

# Answer by turtle_in_mind (score 1)


# 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.

## Answer by turtle_in_mind (score 1)

https://quant.stackexchange.com/a/44203

I’m not sure I understand your question. In the limit as n goes to infinity, the binomial price approaches the continuous price. Your answer is simply k * N(-d2) where N(.) is the CDF of a standard normal with parameters given by a standard put option in BS framework

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.