How to Graph integrals?

Hello everybody,

I'm trying to get this graph:

Using this code:

function res = fun(bs,gamma)
  fun_ode = @(b,z)[(1-gamma)*z(1)/(z(2)-b);z(2)/(z(1)-b);z(1)];
  z0=[1 1 0];
  [B,Z] = ode45(fun_ode,[bs,0],z0);
  res = bs - 0.5*(1+gamma*Z(end,3));
end


gamma = 0.3;
bs = 0.542 
fun_ode = @(b,z)[(1-gamma)*z(1)/(z(2)-b);z(2)/(z(1)-b);z(1)];
z0=[1 1 0];
[B,Z] = ode45(fun_ode,[bs,0],z0);
hold on
plot(B,Z(:,1),'r')
plot(B,Z(:,2),'b')
hold off
grid on

Error using fplot>singleFplot (line 258)
Input must be a function or functions of a single variable.

Error in fplot>@(f)singleFplot(cax,{f},limits,extraOpts,args) (line 218)
hObj = cellfun(@(f) singleFplot(cax,{f},limits,extraOpts,args),fn{1},'UniformOutput',false);

Error in fplot>vectorizeFplot (line 218)
hObj = cellfun(@(f) singleFplot(cax,{f},limits,extraOpts,args),fn{1},'UniformOutput',false);

Error in fplot (line 184)
hObj = vectorizeFplot(cax,fn,limits,extraOpts,args);

But I didn't get it. I would appreciate it if I could get the graph.

Answer 1

[Torsten]

According to proposition 4, the red curve should be the inverse bidding function X and the blue curve should be the inverse bidding function Y for the given value of gamma.

gamma = 0.3;
bs0 = 0.3;
bs = fsolve(@(b)fun(b,gamma),bs0,optimset('Display','None'))

bs = 0.4509

fun_ode = @(b,z)[(1-gamma)*z(1)/(z(2)-b);z(2)/(z(1)-b);z(1)];
z0=[1 1 0];
[B,Z] = ode45(fun_ode,[bs,0],z0);
hold on
plot(B,Z(:,1),'r')
plot(B,Z(:,2),'b')
hold off
grid on

function res = fun(bs,gamma)
  fun_ode = @(b,z)[(1-gamma)*z(1)/(z(2)-b);z(2)/(z(1)-b);z(1)];
  z0=[1 1 0];
  [B,Z] = ode45(fun_ode,[bs,0],z0);
  res = bs - 0.5*(1+gamma*Z(end,3));
end

[Researcher DSGE]

Thanks a lot!

Answer 2

[Walter Roberson]

b is not defined.

This is the same problem as you had for your previous iteration of this question (which you have deleted.) I pointed out the problem there.

[Researcher DSGE]

Dear Walter,

The graphing itself is quite simple, the issue is to get the x and y coordinates for the bidders' values and bids.

I remember that it was necessary to numerically approximate values that fulfill the equations in Proposition 4 for it to be solved. I do not remember the specific procedure I used (only that it was relatively simple and sufficient to reach convergence).

(the integral in (8) has a typo, it is an integral over the interval 0, b(upper bar))

I am sorry that I cannot help you further at this time.

[Walter Roberson]

syms v a 

gamma=0.3;
b_hat=0.542;
fun = v/2;
B(v) = int(fun, v, 0, a)

B(v) = 

fplot(B,[0 1])

Error using fplot>singleFplot (line 258)
Input must be a function or functions of a single variable.

Error in fplot>@(f)singleFplot(cax,{f},limits,extraOpts,args) (line 218)
hObj = cellfun(@(f) singleFplot(cax,{f},limits,extraOpts,args),fn{1},'UniformOutput',false);

Error in fplot>vectorizeFplot (line 218)
hObj = cellfun(@(f) singleFplot(cax,{f},limits,extraOpts,args),fn{1},'UniformOutput',false);

Error in fplot (line 184)
hObj = vectorizeFplot(cax,fn,limits,extraOpts,args);

You are defining B to be a function of v. But you are integrating over v with definite integral bounds that are independent of v, so the result will be independent of v. As a result, B(v) is independent of v, and is instead dependent on the constant a

[Torsten]

@Researcher DSGE

I don't understand the relation between your graphs and Proposition 4. What is v ? Why is b plotted as y-axis and not as x-axis for it seems to be the independent variable ?