Hi,Would appreciate help with this problem.Given q,r,s>0. Does there exist real numbers X and Y such that the following inequality is true?exp(4*(q*X+r*Y))*[erf(X+q)+erf((-q*x-r*Y-s^2)/s)] + exp(4*r*Y)*[erf(-X+q)+erf(Y+r)] + erf(-Y+r) + erf((q*X+r*Y-s^2)/s) < 0Where erf() is the error function.I know that if one chooses s to be sufficiently larger than r and q then the inequaltiy can be solved quite easily. However, I want to solve the inequality for Xand Y (or show the existence of a solution) given that q,r,s are all very similar in magnitude. Although you can feel free to let s>q>rs>r>qr>s>qr>q>sq>r>sq>s>rThanks in advance

Just a quick observation:None of the terms in the objective function are ever < 0. (i.e. erf is never < 0, exp is never < 0)So if you want real numbers I think we're stuck...

Hi,I believe erf can be less than zero (it is an odd function). See http://mathworld.wolfram.com/Erf.html for a description. Thanks for having a look. It'll be great if you can solve it

I guess x=X in your Q?It looks like a question about 'normed' prices:Lx := erf(x/a+1/2*a)*exp(x)+erf(-x/a+1/2*a)*exp(-x),Ly := exp(y)*erf(y/b+1/2*b)+exp(-y)*erf(-y/b+1/2*b),Lz := exp(-z)*erf(1/c*z-1/2*c)-exp(z)*erf(1/c*z+1/2*c)gives me exp(-y)*Lx + exp(-x)*Ly + Lz = yours for X=x/q/2, Y=-y/r/2, z=x-y, a=2*q, b=2*r, c=2*sIn some sense it vaguely reminds me to a triangle inequality.Playing with it (no, I have no solution) I think less 0 needs at least =0(by looking at infinity for the X and Y), right? One way is to look atquotients (which may be easier for those transcendet functions) and thencompare to 1, (exp(-y)*Lx + exp(-x)*Ly) / Lz ...The last may be, not to use x+-y but a multiplicative lookthrough y= k*x or log (caring for signs).What is your solution for large s?

Thanks for the reply AVt. I quite like your way of expressing the inequality. I believe that the equation tends to 0 in a couple of asymptotic case (X,Y ->infinity) but does not go < 0. I ended up thinking more about what you wrote last night and will think more about it (I like the way that Lx and Ly are always positve for all x and y, and that Lz is always negative. This simplifies things quite alot.). There is computational evidence to show that the inequality can be satisfied if s>q+r. I can show this by substituting arbitrary values into the inequality for q,r,s,Y and then plotting a graph of my (or your) equation against X. It turns out that if I let s=q+r then, for whatever Y I choose, there is an X where the graph just touches zero. If I let s>q+r, then I get my inequality to be satisfied for a specific X.A more rigorous (sufficient but not necessary) condition for the inequality to be satisfied is to let X=Y=0 in my equation. Then to choose q,r,s such that the following is trueerf(s) > erf(q)+erf(r) (*)This can clearly be satisfied assuming that erf(q)+erf(r) <1 and then choosing s to be large enough.That is to say that I can show that a sufficient, but not necessary condition for my inequality to be true is for (*) to be satisfied. Also I have an indication that for s>q+r there exists X and Y such that the inequality is true (by my computational arguements). I'm actually looking for a solution which doesn't reuire such constraints on q,r,s. To be honest I don't think it exists, but I'm not completely sure and it would be great if it did.

Hm, I would like to change notation, Lz with negative sign to be positive,so my:=exp(-y)*Lx + exp(-x)*Ly - Lz = yours. I think you reach 0 for large x,y.The function as fct in c is seen to be monotonous by differentiating w.r.t.c (like vega), it is -2/Pi^(1/2)*exp(-1/c^2*(x-y)^2-1/4*c^2).Now one can kick off c by taking limit c = 0 from the right (it is positive),lim_c_0:='limit(my,c=0,right)', for which you consider x < y and y < x separatelyI get that as L0 - 2*exp( abs(x-y)), L0 := (erf(y/b+1/2*b)+1)*exp(y-x)+(erf(x/a+1/2*a)+1)*exp(x-y)+(-erf(x/a-1/2*a)-erf(y/b-1/2*b))*exp(-y-x).Resorting to the error function shows and grouping for a and b shows it to besimplier, not only c vanished, even two erf terms are gone.Hope I did not make to much errors through that and you can continue.

Thanks again for the help AVt (don't think you made any mistakes). I've been working on it a bit and found some results (no solution though). I'll post it at the beginning of next week if I still have any questions, and the results I get by using your ideas (it's too late at the moment).

I find it difficult to say, what should be understood as a solution (andformally it is a semi-analytic set of codim = 1 in the space for the x,yand parameters), may be it depends what you want to 'see'.For that I prefer to look at fct=0, but for 5 variables probably the bestway is to look at limiting cases and find some qualitative discription formoving along the variables from there.lim_c_0 is already stated and lim_c_inf = limit(my, c=infinity) can alsobe given.Now one observes that this is increasing w.r.t. a and also w.r.t. b (asdifferentiating gives an exp term, so it is positive), so the smallestpossible value is for a=b=0, c=infinity (in a limiting sense) and that is-2*exp(-abs(x)-abs(y)), where one has to add 8*sinh(x)*sinh(y) if both xand y are less than 0 (that is just the general case for your X=0=Y).This is always negative - except x,y < 0, where the sinh may quickly addto get s.th. positive. Note that this is symmetric in x and y, but it isindependent of a and b.For the other extremes let a and b reach infinity in lim_c_inf (which isto say: c such that 'my' is a small as posssible, but now let the otherparameters a and b grow). This gives 2*exp(-y-x), hence always positive.Between these extreme case it depends on the concrete parameters andthere i would look at the surface given by x,y.The boundary fct(x,y) = 0 means to intersect the surface with a plane(and at one of its sides your inequality then holds over the reals).That intersection is a complex curve (i.e. dim=1), but as we are overthe reals it may consist of several real curves or additional isolatedpoints (or may be empty). I have not tried to find out more ...Just a final remark: as the fct is non-elementary a solution of fct(x,y)=0for x or y will in general give nothing in terms of 'known' functions Iwould guess, and thus you will have to live with numerical solutions.BTW: what is the origin of your problem?

Hi,I'd also found similar asymptotic limits that you stated. I agree that one needs to look in the intermediate region. Given that problem is 5 dimensional this is hard. We know that the LHS of our inequality is monotonous when we move a,b,or c. I.e. the LHS increases when we increase a and/or b. Also it decreases when we increase c. So maybe we can try the following. Let W:=(LHS:a=b=c). Then solve diff(W,c)=0 for c. This will give us a value for c (in terms of x and y) where our inequality is a maximum or minimum (we want the minimum). I did this but just realised that I made a mistake (hence the edit). I'll check it now.Using this value for c (assuming it exists and that c minimizes W), we can try and solve W<0 for x and y. If this were possible things would be great. Btw the origin of this problem is to do with pricing exotic options. Somewhere down the line it has become essential for me to investigate this inequality. If you can solve the problem that would be great and I'd def have to chat with you. Really appreciate your input thus far.

Actually the above sounds pretty hard so may just have to write some code to slug through all possibilities for W.

Hi,It is too late today for me to see, why a=b=c will help (my feeling is, that istoo symmetric, similar to the boundary situations) ... After some reformulationit means you want to solve exp(-v*c-u^2)+exp(-u*c-v^2)-exp(-(-u+v)^2) = 0 for c,u=x/c, v=y/c (or similar ... = 1).Playing with Maple for v = n/m*u (small integers), this has a solution involvinglog of polynomial roots, but in general I do not see a non-numerical way :-(If it is from exotics my feeling says vola smile is likely treated quite roughlythis way ...

I'll check your differentiation. I too used maple but had forgotten that X and Y are dependent on a and b respectively. So when I set a=b=c and differentiated w.r.t to c I wasn't doing it properly (hence why I editted the last post).Anyway I thought I'd update and say that I wrote a simple C++ code that can scans through values of X,Y,c, where (a=b=c) and returns the minimum value. It appears that this minimum value is never negative. I already have strong evidence that if c>a+b, then there does exist X and Y s.t the original inequality is satisfied, however, I'm close to having pursuaded myself that for c<a+b the inequality cannot be satisfied. Maybe I can rewrite the original inequality with a=b and c=3*a/2. This way c>a,b and c<a+b. I suppose there may be a chance that the a=b condition is too symmetric and that a null result wont have proven anything, however, it could be worth a try. Btw I haven't thought much about what I just said but will do a bit more now.

The original inequality holds iff the transformed holds (since you thinkthe parameters as being given), so I think there is no need to care fordependency, no?

For a=b=c there is seems there is no minimum for finite c: plottingsuggests a minimum would need x=y, solving diff(W,c)=0 for c gives2 solutions plus a restriction -oo < x < ln(2) to be real. Since cis positive one looks at x (=y) in -oo < x < 0 and 0 < x < ln(2)choosing the appropriate solution for c. The minimum is x=0 (other-wise everthing is positive), which needs c=0.I googled for "analytic inequalities", but can not see, how yourproblem fits into "superadditive" or "subadditive". But may be thewords may helps you to find out more. It will be non-trivial (andnot 'only' a bivariate variant) to apply findings as long as youcan not restrict your parameters (=volatilities?). My gut feelingis your setting is a bit too general, may be you need assumptionswhich are close to reality and carefully test solutions for theirvalidity to go further (say: specific models) - as soon as you dobring in stock dynamics your general solution (if ever exists) hasa chance to be non-stable.Or in plain words: I give up.

Thanks for the help. I'll def google those key words. I'm trying now to look at specific cases and playing with program. Thanks for the help on the minima aswell.