# Problem with the concept of Dollar Gamma

I was reading up on variance swaps and encounter the notion of Dollar Gamma, which is defined as the change in dollar value of the Dollar Delta (Δ * S) for a 1% change in spot (S). The formula for Dollar Gamma is then given as $Γ = Γ/100 * S^2. However, as I tried to prove it, I ran into an issue. The proof is below.$Δ = Δ * S

$Δ' = Δ' * S' (Function 1)$Δ' = (Δ + Γ * S/100) * (S + S/100)

\$Δ' = Δ * S + Γ/100 * S^2 + [1% * (Δ * S + Γ/100 * S^2)] (Function 2)

Clearly from Function 2, the change in Dollar Delta is not only the Dollar Gamma as per definition but also the part in the square brackets. However, if in Function 1, the new Delta is multiplied by S instead of S', then the definition would be correct. So my question is whether the new Dollar Delta is calculated by multiplying the new Delta (Δ') with the previous spot (S) or my proof in Function 2 is correct but in practice traders just ignore the terms in the square brackets because they're negligible?

Please forgive any formatting errors for the proof.

Thanks.

The answer to the first question is to take the number of shares that your delta changes by, times the spot. It’s important not to take the difference in cash delta i.e. do not take something like $$\Delta_+S_+ - \Delta_-S_-$$ because this only represents a monetary risk scaling due to price fluctuations, not a convexity risk: a simple stock has no gamma, but this formula would show you a change in cash delta when spot changes which in itself is perfectly reasonable (you have more dollar risk on your holding when spot is high) but is not indicative of a gamma risk.
$$PL_{\Gamma_{\}}\approx \frac{1}{2}\frac{\Gamma_{\}}{1\%}.(\frac{dS}{S})^2$$