Could someone please give me an evidence (or where to find an evidence) if (and if so, then why) prices of an American Call on a dividend paying underlying are convex in the strike?!I've browsed the internet and found a few hints, but not a real prove.What I found is that "it is well known" that prices of American options are convex in the underlying price (gamma >= 0) and that the optimal stopping boundary is a convex function for American Puts.Can you say that the price of an American Call on a dividend paying underlying is convex in strike because it is the discounted expectancy of a convex payoff-function (S-K)+ with a convex stopping time?Many thanks in advance.

Just thinking out loud, here is some intuitive reasoning for it. As I recall, it's pretty easy to establish that, for European options with general payoffs g(S_T), the valuation function V(S)is convex in S if g(.) is convex. Then, one can think of the American call as determined by a backwards recursion ofsmall time steps dt. As each time step you are evaluating C(S) = e^(-r dt) E[C*(S + dt)], where C* is either a continuationvalue or an exercise value (assume the optimal exercise boundary is known). This C* is still a convex function of S (picture it), so each time step is simply a mini-euro style evaluation. Hence, C(S) must be convex in S again. By induction, this convexity holds all the way back to any point in time, and by homogeneity in S and K, C_SS = C_KK.regards,

If the payoff function h(K,S) is convex in K, thenh(a*K_1 + (1-a)*K_2, S) <= a*h(K_1,S) + (1-a)*h(K_2,S).Then taking expected value under risk neutral measure, using monotonicity of expected value and linearityE[ B_t * h(a*K_1 + (1-a)*K_2,S_t) ] <= a * E[ B_t * h(K_1,S_t)] + (1-a) * E [ B_t * h(K_2,S_t) ]Now take maximum over t = all possible stopping times. Let the right hand side be written as A_t + B_t, then notemax_t (A_t + B_t) <= (max_t A_t) + (max_t B_t), and we can pull the constant a and (1-a) terms out, and the result follows.You mentioned a convex stopping time, what do you mean by this ? Is there something above that requires an extra condition on the stopping time that I am missing?Regards.

Thank you both for your replies.@AlanRegarding your last point, C_SS = C_KK; this is not true for European Calls, so why should it be for American Calls?For European:C_SS (=gamma) = N(d1)*exp(-rT)/(S*sigma*sqrt(T))C_KK = N(d2)*exp(-rT)/(K*sigma*sqrt(T))Both are > 0, which means that the price is convex in S and K, but they are not equal.@aprokopiwInteresting and easy way, but I don't get what A_t and B_t are.According to my professor the convex stopping time comes into play at the point where you take the maximum.

Sorry, I meant to say S^2 C_SS = K^2 C_KK

I think I made a typo when I said left hand side, I meant right hand side. I was saying,Let A_t = a * E[ B_t * h(K_1,S_t)]B_t = (1-a) * E [ B_t * h(K_2,S_t) ]I had thatE[ B_t * h(a*K_1 + (1-a)*K_2,S_t) ] <= A_t + B_t.Now taking the max over all {t a stopping time}, the left hand side is the price of the american call at strike a*K_1 + (1-a)*K_2, the right hand side ismax (A_t + B_t), but I am claiming max (A_t + B_t) <= max A_t + max B_t , and then the right hand side just becomes the affine sum of the two american calls with price K_1 and K_2.You are saying you need some condition on the stopping time, but I don't see this yet.Just consider any indexing set J, and suppose to each j in J we have numbers A_j and B_j.Then surely a*A_j <= max_j(a*A_j) for all j, and (1-a)*B_j <= max_j((1-a)*B_j) for all j, so that a*A_j + (1-a)*B_j <= max_j (a*A_j) + max_j((1-a)*B_j) for all j. Hencemax_j (a*A_j + (1-a)*B_j) <= max_j (a*A_j) + max_j((1-a)*B_j) = a* max_j(A_j) + (1-a) * max_j(B_j).In other words, the max_j ( ) function is already convex.Now take J = set of all stopping times, and apply it to the above, so I still don't see what is meant by a convex stopping time.

Ah, I think I got it this time.Your B_t has two meanings, first the discounting function and second an abbreviation for the second term on the right hand side. Got a bit confused about this first.Looks like you're right about the stopping time. This would mean that the proof of the convexity is much simpler and more straightforward than I thought.@AlanI don't know if I am overlooking something obvious but I don't see why S^2*C_SS = K^2*C_KK.For an European Call this would mean:S^2*C_SS = S*N(d1)*exp(-rT)/(sigma*sqrt(T))K^2*C_KK = K*N(d2)*exp(-rT)/(sigma*sqrt(T))=> S*N(d1) = K*N(d2).Perhaps this is true, but if so I didn't know about it.Nevertheless I hope the proof of aprokopiw stands all forthcoming questions.Thank you very much both of you.