Hi Everyone,Title says my question. Suppose you have a european call option that we price using BS formula. How can we prove that its price is increasing in volatility (sigma)? I do know the intution behind it so please don't make intuitive arguments. I want to find out how to prove it. Thanks a lot!

QuoteOriginally posted by: bluehonourHi Everyone,Title says my question. Suppose you have a european call option that we price using BS formula. How can we prove that its price is increasing in volatility (sigma)? I do know the intution behind it so please don't make intuitive arguments. I want to find out how to prove it. Thanks a lot!Because vega is non-negative.

Suppose you are asked this in an a job interview. Will you remember the formula for vega of Euro call? Will you just say its vega is non negative? How will you respond if the interviewer asks "prove that vega is nonnegative"? This is a real interview question, by the way.

QuoteOriginally posted by: bluehonourSuppose you are asked this in an a job interview. Will you remember the formula for vega of Euro call? Will you just say its vega is non negative? How will you respond if the interviewer asks "prove that vega is nonnegative"? This is a real interview question, by the way.Without the derivative of the vega, a rough proof can be ? d1 increases as sigma increases (linear)d1 >= d2 N(d1) >= N(d2) ( CDF is monotone function)Ok, C = SN(d1) - ke^-rT N(d2)small assumption of C >= 0 now two terms of call price are monotone increasing in sigma(hand waving)... => C increases as sigma increases, the only other case is the trivial case.what do u guys say, can make more rigourous maybe ? Why would anyone ask you to derive the derivative of call price with sigma in a interview ? Doesnt sound too fruitful a question K.

To establish the result you'll also need to require convexity of the option payoff. Then you can invoke Jensen's inequality

QuoteOriginally posted by: bluehonourSuppose you are asked this in an a job interview. Will you remember the formula for vega of Euro call? Will you just say its vega is non negative? How will you respond if the interviewer asks "prove that vega is nonnegative"? This is a real interview question, by the way.Yes, I memorize the formula of vega. Even if you cannot memorize it, you should be able to derive it from the BS formula, which is again something you should memorize before an interview. I think everyone interested in option pricing should derive greeks on himself at least once in his life.

for informal proof, take an example. say call option. Now price=E[max(s-k,0)] as vol increases higher payoffs get included(as optionprice is a weighted average, weights being probability of correspondng payoffs) with higher probability because normal curve is more spread. for formal proof use vega or define probablity measur dp/dx=(1/sqrt(2*pi*variance))/exp( -0.5*(x-m)^2 / variance) so p beingan increasing function of vol(variance) it follows... its hard to use jensen's inequality directly.

Athletico, Maybe I am missing something, But if we can assume convexity of call price w.r.t to sigma then that means the first derivative is non decreasing which defeats the purpose no ? This is what I think you have in mind, Let d1' = d1 + epsilon & d2' = d2 + epsilon then d1 > d1' > d2 > d2' we have to prove N(d1) - N(d1')/(d1 - d1') >= N(d2) - N(d2')/ d2 - d2'And this I think requires convexity.Macca9,I think the approach of expressing call price generically in terms of a up and down move can get us into trouble. Since I dont know how you plan to avoid using the max(S-K,0) function ? K.

>> But if we can assume convexity of call price w.r.t to sigmaSorry - I meant payoff convexity with respect to underlying. This is clearly the case for vanilla options. Then you can take an inductive argument tack that starts with Jensen to establish the basis condition: Any underlying variance > 0 increases the value of a Euro call relative to the zero variance case.

To expand on Athletico's argument:We need to show E(f(XY)) >= E(f(X))when X,Y independent and E(Y)=1E(f(XY)=E_X(E_Y(f(XY))by Jensen's inequalityE_Y(f(XY)) >=f(E_Y(XY)) = f(X)and we are done.

QuoteOriginally posted by: mjTo expand on Athletico's argument:We need to show E(f(XY)) >= E(f(X))when X,Y independent and E(Y)=1E(f(XY)=E_X(E_Y(f(XY))by Jensen's inequalityE_Y(f(XY)) >=f(E_Y(XY)) = f(X)and we are done.mj and Athletico,sorry but i am missing it. Y is vol ?I thought you would take it to be const right.. u are assuming its a r.v ?a little help thanks,K.

we write S as a product of two lognormalsX is the one with lower vol, Y is the extra vol.i.e.X = S_0 exp((r-0.5 sigma^2) T + + sigma \sqrt(T) Z_0)Y = exp( -0.5 nu^2 T +nu \sqrt(T) Z_1)where Z_j are independent N(0,1) and \nu expresses the extra variance

ofcourse that make sense.thanks, mj