When I assess tail hedges for high-flyers like NVIDIA (NVDA), I don’t start with the options chain. I start by making the crash scenario explicit. Only then do I ask whether the listed puts offer positive expected value under my view of the world. This post walks through the exact quantitative workflow I use, entirely in maths–no custom code required.

The key idea: separate

  1. A real-world model for NVDA’s long-term behaviour and “AI bubble burst” probability,
  2. A subjective crash/no-crash scenario set for terminal prices,
  3. A simple expected-value check of listed puts against that model.

1. Define the State Variables

Fix a valuation date $t_0$. On 10 Nov 2025, NVDA closed at:

\[I_0 = \$196.14.\]

From the option chain, the market priced two-year risk with an at-the-money implied volatility of

\[\sigma = 42.65\%\quad (\text{annualised}).\]

Because we’re judging the risk of an “AI bubble” bursting, we must translate that into a price event. I define a bubble burst as a drawdown of at least 53% from today’s level:

\[K = (1 - 0.53),I_0 = 0.47,I_0 \approx \$92.19.\]

So the “bubble burst by two years” event is:

\[{\text{bubble burst by 2y}} := { I_2 < K }.\]

Finally, I specify a median two-year price that reflects my base case, $M = 268$. This is not the mean; it’s the level I think is “most typical” if NVDA avoids a full bubble collapse. Using the median rather than the mean avoids some lognormal skew issues and gives a clean way to back out a real-world drift $\mu$.

2. Calibrate the Lognormal Process

I model NVDA under the real-world (physical) measure as a geometric Brownian motion (GBM):

\[\frac{dI_t}{I_t} = \mu dt + \sigma dW_t,\]

where

  • $\mu$ is the real-world expected return (what I think NVDA earns on average),
  • $\sigma$ is its volatility, here taken from the 2-year ATM implied vol.

Under GBM, the terminal log-price after $T$ years is normal:

\[\ln I_T \sim \mathcal{N}\big(m, v^2\big), \qquad m = \ln I_0 + (\mu - \tfrac{1}{2}\sigma^2)T, \qquad v^2 = \sigma^2 T.\]

The median of a lognormal is

\[\mathrm{median}(I_T) = e^{m - \tfrac{1}{2}v^2}.\]

Plug in the GBM parameters:

\[\mathrm{median}(I_T) = I_0 \exp\big[(\mu - \tfrac{1}{2}\sigma^2)T\big].\]

I want $\mathrm{median}(I_2) = M$. Solving for $\mu$ yields

\[\mu = \frac{\ln(M / I_0)}{T} + \tfrac{1}{2}\sigma^2.\]

With $I_0 = 196.14$, $M = 268$, $T = 2$ and $\sigma = 42.65\%$,

\[\mu \approx 24.70\%\ \text{per year}.\]

Interpretation: under my base-case assumptions (implied vol, chosen median), I’m implicitly saying NVDA has about a 25% real-world annual return over the next two years if the AI story evolves “normally”.

3. Bubble-Burst Probability From the Lognormal Model

Under this GBM, we can compute the real-world probability that NVDA ends below the crash threshold (K) at horizon $T = 2$ years:

\[P_{\text{burst}}(2) = \mathbb{P}(I_2 < K) = \mathbb{P}(\ln I_2 < \ln K) = \Phi\left(\frac{\ln K - m}{\sigma \sqrt{T}}\right),\]

where $\Phi$ is the standard normal CDF and

\[m = \ln I_0 + (\mu - \tfrac{1}{2}\sigma^2)T.\]

Numerically, this works out to

\[P_{\text{burst}}(2) \approx 3.84\%.\]

So, under this particular calibration, I’m effectively saying:

“There’s about a 3.8% real-world chance that NVDA trades below $92 (a 53% drawdown) in 2 years.”

This is not the market’s risk-neutral probability (the one consistent with option prices); it’s my own real-world view derived from median, vol and spot.

From a Single Horizon to a Hazard Rate

To avoid re-doing the GBM algebra for every maturity, I approximate the time-profile of burst risk with a constant hazard rate $\lambda$:

\[P_{\text{burst}}(T) = 1 - e^{-\lambda T}.\]

Given $P_{\text{burst}}(2)$, we solve

\[\lambda = -\frac{\ln\big(1 - P_{\text{burst}}(2)\big)}{2} \approx 0.0196\ \text{yr}^{-1}.\]

Interpretation:

  • Think of $\lambda$ as a Poisson-style crash intensity.
  • It implies a smooth curve of burst probabilities for all maturities: $P_{\text{burst}}(T) = 1 - e^{-\lambda T}$, which I can then plug into the options analysis for any expiry between 9 and 24 months.

This is a modelling choice: I’m summarising a detailed GBM calculation at 2 years by a simpler hazard process that is easy to extrapolate.

4. A Two-Regime Scenario Mixture for Terminal Prices

The GBM gave me a single number $P_{\text{burst}}(T)$ at each maturity. It did not give a detailed crash distribution. For options, I want something more interpretable than a pure lognormal: a mixture of:

  1. Burst regime: AI bubble has burst by expiry; NVDA ends in a low price band.
  2. No-burst regime: AI bubble hasn’t burst; NVDA remains elevated.

I model NVDA’s terminal price (S_T) as a discrete mixture:

Burst regime: $S_T \mid \text{burst}$

Here I assume a handful of possible post-bubble levels with conditional probabilities that sum to one:

  • $90 (probability 0.35)
  • $80 (0.25)
  • $70 (0.18)
  • $60 (0.13)
  • $50 (0.09)

This says: if the bubble bursts, most of the mass is clustered just below today’s crash threshold, with smaller weight in deeper capitulation.

No-burst regime: $S_T \mid \text{no burst}$

If the bubble does not burst, I assume NVDA remains in a higher band, again with conditional probabilities summing to one:

  • $250 (probability 0.30)
  • $268 (0.50)
  • $300 (0.20)

You can think of $268 as the “central” (median-ish) outcome, with some chance of mild disappointment ($250) or further euphoria ($300).

These probabilities are subjective. The ones above are chosen to be internally consistent (each regime sums to 1) and to reflect a plausible crash/no-crash story. You can change them without touching the rest of the machinery.

Unconditional mixture

Given a maturity $T$, the hazard model gives a burst probability $P_{\text{burst}}(T)$. The unconditional distribution of $S_T$ is:

\[\mathbb{P}(S_T = s) = P_{\text{burst}}(T)\,\mathbb{P}(s \mid \text{burst}) + \big(1 - P_{\text{burst}}(T)\big)\,\mathbb{P}(s \mid \text{no burst}).\]

This replaces the continuous lognormal with a small discrete distribution that is:

  • Easy to reason about in economic terms (actual price levels),
  • Easy to plug into option payoff formulas.

5. Expected Payoff and Profit Per Put

For a European put with strike $K_{\text{put}}$ and premium $\Pi$ (per share), the payoff at expiry is:

\[\text{payoff}(S_T) = \max\big(K_{\text{put}} - S_T, 0\big).\]

Given our discrete unconditional distribution ({(s_i, p_i)}), the expected payoff per share at maturity is:

\[\mathbb{E}[\text{payoff}] = \sum_i p_i,\max(K_{\text{put}} - s_i, 0).\]

If we (mildly) ignore discounting over 1-2 years, the expected profit per share under our view is:

\[\mathbb{E}[\text{profit}] \approx \mathbb{E}[\text{payoff}] - \Pi.\]

In a fully precise setup you’d discount the expected payoff by (e^{-rT}) at the risk-free rate (r), but for shortish maturities and moderate interest rates this is a second-order detail relative to the crash assumptions and premium size.

Example Puts and Premiums

I pulled mid premiums from IBKR for two maturities (all contracts standard 100-share lots):

  • 18 Sep 2026 (about $T = 0.855$ years from 10 Nov 2025): strikes $110-$150, premiums roughly $3.40-$11.28.
  • 17 Sep 2027 (about $T = 1.852$ years): strikes $110-$150, premiums roughly $8.45-$20.10.

For each triplet $(T, K_{\text{put}}, \Pi)$, I:

  1. Compute $P_{\text{burst}}(T) = 1 - e^{-\lambda T}$.
  2. Form the unconditional discrete distribution over $S_T$ using the burst/no-burst mixture.
  3. Compute $\mathbb{E}[\text{payoff}]$ and $\mathbb{E}[\text{profit}]$.

Under my real-world crash view, every contract I looked at had negative expected profit. The “least bad” was the Sep 2026 $110 put:

\[\mathbb{E}[\text{payoff}] \approx \$0.60,\quad \Pi = \$3.40,\quad \mathbb{E}[\text{profit}] \approx -\$2.80.\]

So even the best-looking put, in expectation, loses most of the premium; these are insurance policies, not positive-EV trades, under my assumptions.

A summary of the key rows (per share):

Valuation Expiry $T$ (yrs) Strike Premium $P_{\text{burst}}(T)$ $\mathbb{E}[\text{payoff}]$ $\mathbb{E}[\text{profit}]$
2025-11-10 2026-09-18 0.855 110 3.40 1.66% 0.60 -2.80
2025-11-10 2026-09-18 0.855 120 4.70 1.66% 0.77 -3.93
2025-11-10 2027-09-17 1.852 110 8.45 3.60% 1.30 -7.15
2025-11-10 2027-09-17 1.852 150 20.10 3.60% 2.72 -17.38

(The full table includes all strikes from 110 to 150 for both maturities.)

6. Interpretation & Next Moves

From a finance and probability point of view, the story is:

  • Given:

    • my moderate burst probability ($P_{\text{burst}}(2) \approx 3.8\%$),
    • the depth of my crash scenario (50-75% drawdowns), and
    • the current option premiums,

    NVDA puts look expensive as tail hedges. They have negative expected value under my subjective crash distribution; I’d be paying away premium for insurance I don’t think is fairly priced.

  • That doesn’t mean they’re “wrongly priced” in a risk-neutral sense; it means:

    • under my real-world view, the crash risk doesn’t justify the quoted premium.

To turn a put into a worthwhile bet rather than pure insurance, at least one of the following must change:

  1. Higher burst probability New information (macro, AI fundamentals, positioning) that lifts your real-world crash probability into double digits–e.g. a regime where $P_{\text{burst}}(2)$ is 10-20%, not ~4%.

  2. More brutal crash scenarios You revise your post-burst levels lower e.g. NVDA at \$40 - \$60 instead of \$50 - \$90, increasing $\mathbb{E}[\text{payoff}]$ at a given strike.

  3. Cheaper implied volatility / premiums Implied vol compresses or the market underprices downside skew, so $\Pi$ falls relative to your expected payoff.

Mathematically, the workflow is fully deterministic:

  1. Define a price-based burst event ((K)).
  2. Calibrate a real-world GBM from spot, implied vol, and your median target.
  3. Extract $P_{\text{burst}}(T)$ and convert to a hazard $\lambda$.
  4. Build a two-regime price mixture (burst/no-burst) that reflects your narrative.
  5. Compute expected payoff and profit for each listed put.

If NVDA’s environment changes–spot, implied vol, fundamental outlook–you just update:

  • $I_0$,
  • $\sigma$,
  • $M$,
  • and your burst threshold (K),

and re-run the same chain of calculations. The method itself doesn’t care whether the ticker is NVDA, an AI ETF, or some other “bubble candidate”; it’s a general way to make your crash view and option selection mathematically consistent.

If you want to recreate every number without re-reading the prose, the appendix below walks through each calculation step by step.

Appendix - Detailed Calculations

This appendix walks through the main numbers in the post step by step:

  1. Calibrating the lognormal (GBM) model
  2. Computing the 2-year bubble-burst probability
  3. Converting that to a hazard rate
  4. Example expected payoff calculation for a put under the burst/no-burst mixture

Feel free to hide this from casual readers if it’s too nerdy; it’s mainly for people who want to see the math all the way down.

A. Calibrating the Lognormal Model

We start with:

  • Current NVDA proxy level (treat it as an “AI index”): $I_0 = 196.14$
  • Two-year horizon: $T = 2$
  • Annual volatility from 2-year ATM implied vol: $\sigma = 42.65\% = 0.4265$
  • Chosen median 2-year level: $M = 268$

Under geometric Brownian motion:

\[\frac{dI_t}{I_t} = \mu dt + \sigma dW_t,\]

we have:

\[\ln I_T \sim \mathcal{N}\big(m,\ v^2\big), \quad m = \ln I_0 + (\mu - \tfrac{1}{2}\sigma^2)T, \quad v^2 = \sigma^2 T.\]

The median is:

\[\mathrm{median}(I_T) = I_0 \exp\big[(\mu - \tfrac{1}{2}\sigma^2)T\big].\]

We want $\mathrm{median}(I_2) = M$, so solve for $\mu$:

\[\mu = \frac{\ln(M / I_0)}{T} + \tfrac{1}{2}\sigma^2.\]

Now plug in the numbers.

  1. Ratio $M / I_0$:
\[\frac{M}{I_0} = \frac{268}{196.14} \approx 1.3659.\]
  1. Log of that ratio:
\[\ln\left(\frac{M}{I_0}\right) \approx \ln(1.3659) \approx 0.31216.\]
  1. Half the variance:
\[\sigma^2 = 0.4265^2 \approx 0.1819, \quad \tfrac{1}{2}\sigma^2 \approx 0.09095.\]
  1. Drift $\mu$:
\[\mu = \frac{0.31216}{2} + 0.09095 \approx 0.15608 + 0.09095 \approx 0.24703.\]

So the implied real-world drift is:

\[\mu \approx 24.70\% \ \text{per year}.\]

This is the annualised growth rate that makes $M = 268$ the median 2-year outcome, given volatility of 42.65%.

B. Probability NVDA Ends Below the Crash Threshold in 2 Years

We define the crash threshold:

\[K = 0.47 \times I_0 = 0.47 \times 196.14 \approx 92.1858.\]

This corresponds to roughly a 53% drawdown.

We want:

\[P_{\text{burst}}(2) = \mathbb{P}(I_2 < K).\]

Under GBM:

\[\ln I_2 \sim \mathcal{N}(m, v^2), \quad m = \ln I_0 + (\mu - \tfrac{1}{2}\sigma^2)T, \quad v = \sigma\sqrt{T}.\]
  1. Compute $m$.

We already have:

  • $\ln I_0 \approx \ln(196.14) \approx 5.27883$
  • $\mu - \tfrac{1}{2}\sigma^2 \approx 0.24703 - 0.09095 \approx 0.15608$

So:

\[m = 5.27883 + 0.15608 \times 2 \approx 5.27883 + 0.31216 \approx 5.59099.\]
  1. Compute $v$:
\[v = \sigma\sqrt{T} = 0.4265 \times \sqrt{2} \approx 0.4265 \times 1.41421 \approx 0.60316.\]
  1. Standardise $\ln K$.

First:

\[\ln K \approx \ln(92.1858) \approx 4.52403.\]

Then compute the z-score:

\[z = \frac{\ln K - m}{v} = \frac{4.52403 - 5.59099}{0.60316} = \frac{-1.06696}{0.60316} \approx -1.76931.\]
  1. Convert z to probability using the standard normal CDF $\Phi$:
\[P_{\text{burst}}(2) = \mathbb{P}(\ln I_2 < \ln K) = \Phi(z) \approx \Phi(-1.76931) \approx 0.03842.\]

So:

\[\boxed{P_{\text{burst}}(2) \approx 3.84\%}\]

Under this calibration, there’s about a 3.8% real-world chance NVDA ends below $92 in 2 years.

C. Converting the 2-Year Probability to a Hazard Rate

To get a smooth time profile of burst probabilities, we approximate with a constant hazard rate $\lambda$ such that:

\[P_{\text{burst}}(T) = 1 - e^{-\lambda T}.\]

At $T = 2$, we know $P_{\text{burst}}(2) \approx 0.038421$. Solve for $\lambda$:

\[1 - e^{-2\lambda} = 0.038421 \quad\Rightarrow\quad e^{-2\lambda} = 1 - 0.038421 = 0.961579.\]

Take logs:

\[-2\lambda = \ln(0.961579) \quad\Rightarrow\quad \lambda = -\frac{\ln(0.961579)}{2}.\]

Numerically:

  • $\ln(0.961579) \approx -0.03918$
  • So: $ \lambda \approx -\frac{-0.03918}{2} \approx 0.01959. $

Thus the constant hazard rate is:

\[\boxed{\lambda \approx 1.96\% \ \text{per year}}.\]

You can then compute the implied burst probability at any horizon $T$ via:

\[P_{\text{burst}}(T) = 1 - e^{-\lambda T}.\]

Examples:

  • For $T = 0.855$ years ($\approx$ Sep 2026 expiry): $P_{\text{burst}}(0.855) = 1 - e^{-0.01959 \times 0.855} \approx 1 - e^{-0.01677} \approx 1 - 0.98338 \approx 0.01661 = 1.661\%$.

  • For $T = 1.852$ years ($\approx$ Sep 2027 expiry): $P_{\text{burst}}(1.852) = 1 - e^{-0.01959 \times 1.852} \approx 1 - e^{-0.03630} \approx 1 - 0.96436 \approx 0.03563 = 3.563\%$.

These match the rough 1.66% and 3.56% figures used in the main text.

D. Example: Expected Payoff of a Put Under the Burst/No-Burst Mixture

To evaluate a put, we use the discrete mixture of burst and no-burst regimes.

1. Scenario setup

Burst regime (conditional on bubble burst by expiry):

Price (s) Conditional prob
90 0.35
80 0.25
70 0.18
60 0.13
50 0.09

No-burst regime (conditional on no bubble burst by expiry):

Price (s) Conditional prob
250 0.30
268 0.50
300 0.20

For a given maturity $T$, the unconditional probability of each price is:

\[\mathbb{P}(S_T = s) = P_{\text{burst}}(T) \cdot \mathbb{P}(s \mid \text{burst}) + \big(1 - P_{\text{burst}}(T)\big) \cdot \mathbb{P}(s \mid \text{no burst}).\]

2. Example: 2-Year-ish expiry, strike $K_{\text{put}} = 110$

Take the Sep 2027 option with:

  • Horizon $T \approx 1.852$ years,
  • Strike $K_{\text{put}} = 110$,
  • Market premium (per share) $\Pi = 8.45$ (example from IBKR),
  • $P_{\text{burst}}(T) \approx 3.563\%$ from the hazard model.

Let:

  • $p_b = P_{\text{burst}}(T) \approx 0.03563$,
  • $p_{nb} = 1 - p_b \approx 0.96437$.

Expected payoff per share:

\[\mathbb{E}[\text{payoff}] = \sum_{\text{burst}} p_b \, p(s \mid \text{burst})\,\max(110 - s, 0) + \sum_{\text{no-burst}} p_{nb}\, p(s \mid \text{no burst})\,\max(110 - s, 0).\]

Compute each term.

Burst regime payoff contributions:

  • At $s = 90$: payoff $= \max(110-90,0)=20$; contribution $0.03563 \times 0.35 \times 20 = 0.03563 \times 7 \approx 0.2494$.

  • At $s = 80$: payoff $=30$; contribution $0.03563 \times 0.25 \times 30 = 0.03563 \times 7.5 \approx 0.2672$.

  • At $s = 70$: payoff $=40$; contribution $0.03563 \times 0.18 \times 40 = 0.03563 \times 7.2 \approx 0.2565$.

  • At $s = 60$: payoff $=50$; contribution $0.03563 \times 0.13 \times 50 = 0.03563 \times 6.5 \approx 0.2316$.

  • At $s = 50$: payoff $=60$; contribution $0.03563 \times 0.09 \times 60 = 0.03563 \times 5.4 \approx 0.1924$.

Sum of burst contributions:

\[0.2494 + 0.2672 + 0.2565 + 0.2316 + 0.1924 \approx 1.1971.\]

No-burst regime payoff contributions:

At 250, 268, 300 the payoff is zero:

  • $\max(110 - 250, 0) = 0$
  • $\max(110 - 268, 0) = 0$
  • $\max(110 - 300, 0) = 0$

So all no-burst contributions are zero.

Therefore, total expected payoff:

\[\boxed{\mathbb{E}[\text{payoff}] \approx 1.20 \ \text{per share}}.\]

With a premium $\Pi = 8.45$, the expected profit per share is:

\[\mathbb{E}[\text{profit}] \approx 1.20 - 8.45 = -7.25 \ \text{per share}.\]

(Using precise machine arithmetic gives $\mathbb{E}[\text{payoff}] \approx 1.1971$ and $\mathbb{E}[\text{profit}] \approx -7.25$.)

For one standard contract (100 shares):

\[\mathbb{E}[\text{profit per contract}] \approx -\$725.\]

This is why, under the specific crash probabilities and scenarios I picked, the Sep 2027 $110 put behaves more like insurance (negative expected value) than a positive-EV speculative short, even though it pays out nicely in the crash cases.