A call and a stop-protected perp have the same job.
Cap downside to get asymmetric upside.
How do they compare in doing their job?
Since a perpetual is a trade entered at the current price, we choose a near-the-money call option (85k, spot ~84.2k).
We match leverage by calculating the option's omega (cash delta / premium), which puts the stop 1/Ω below entry, about 5% here.
Then we calculate the number of perpetual contracts such that max risk is the same:
Q × (Entry − Stop) = Premium
We simulate prices assuming:
• 20% drift (you're long for a reason)
• 8% funding
• 32.2% realized vol
• 34.7% implied vol
Results.
Call EV: +$737. Perp + stop EV: +$694.
First observation: the expected values are about the same.
So you pay for asymmetry one way or another: through premium, or through the moves you miss after a stop-out.
Most of the premium cost is earned back through full payoff capture. The part that doesn't roughly matches what stop-outs cost, provided that implied vol isn't excessive compared to future realized vol.
Second observation: the payoff distributions differ.
The left tail is the same by design, since we matched max loss, provided the stop fills. The middle of the distribution favors the perp. The right tail favors the call.
That's logical. You pay option premium to avoid opportunity loss. This pays off if there is a payoff. If price ends near entry, there was little opportunity to lose.
The heart of the matter is that the choice between stop and premium is a bet inside the bet, to be judged on expected volatility.
If implied looks cheap, lean towards the call: realized above implied gives you an edge.
If you expect a steady trend with little chop, i.e. realized below implied, use the perp with a stop.