Defining and plotting 2 parametric functions

I'm new to Matlab, I defined the following functions:

function L1 = Like1(y)
P0 = 1/(2*pi*0.6)*exp(-1/(2*0.36)*(x^2+y^2-2*0.8*x*y));
P1 = 1/(2*pi*1)*exp(-1/(2*1)*(x^2+y^2));
L1 = int(P1,x,-inf,0)/int(P0,x,-inf,0);
end

function L2 = Like2(y)
P0 = 1/(2*pi*0.6)*exp(-1/(2*0.36)*(x^2+y^2-2*0.8*x*y));
P1 = 1/(2*pi*1)*exp(-1/(2*1)*(x^2+y^2));
L2 = int(P1,x,0,+inf)/int(P1,x,0,+inf);
end

function I1 = Int1(y)
P0 = 1/(2*pi*0.6)*exp(-1/(2*0.36)*(x^2+y^2-2*0.8*x*y));
I1 = int(P0,x,-inf,0);
end

function I2 = Int2(y)
P0 = 1/(2*pi*0.6)*exp(-1/(2*0.36)*(x^2+y^2-2*0.8*x*y));
I2 = int(P0,x,0,+inf);
end

function I3 = Int3(y)
P1 = 1/(2*pi*1)*exp(-1/(2*1)*(x^2+y^2));
I1 = int(P1,x,-inf,0);
end

function I4 = Int4(y)
P1 = 1/(2*pi*1)*exp(-1/(2*1)*(x^2+y^2));
I2 = int(P1,x,0,+inf);
end

and I'm trying to plot the following two functions:

function alpha = error1(T)
syms y;    
alpha = int(Int1(y),y,solve(Like1(y)>T,y))+ int(Int2(y),y,solve(Like2(y)>T,y));
end

function beta = error2(T) 
syms y;
beta = int(Int3(y),y,solve(Like1(y)<T)) + int(Int4(y),y,solve(Like2(y)<T))
end

I have 2 questions , when I run this function editer, I keep getting the error "Undefined function or variable 'T'".

Once I solve this issue, what would be the next step to plot error2(T) vs. error1(T).

[Walter Roberson]

Exactly what are you running in the function editor? How are you "running" in the function editor?

[Walter Roberson]

In your functions, x is undefined. The context suggests that you need a

syms x

in each of them.

[Walter Roberson]

In Like2 you have

L2 = int(P1,x,0,+inf)/int(P1,x,0,+inf);

that is dividing something by itself, so the result is always going to be 1. Perhaps you wanted the numerator to be P0 instead of P1.

Question: is it the case that the P0 and P1 in Like1 and Like2 are exactly the same expressions?

[Maggie Mhanna]

Thanks for your answer Walter, I corrected the L2 expression. My function becomes:

function alpha = error1(T)

syms x y

P0 = 1/(2*pi*0.6)*exp(-1/(2*0.36)*(x^2+y^2-2*0.8*x*y));

P1 = 1/(2*pi*1)*exp(-1/(2*1)*(x^2+y^2));

L1 = int(P1,x,-inf,0)/int(P0,x,-inf,0);

L2 = int(P1,x,0,+inf)/int(P0,x,0,+inf);

I1 = int(P0,x,-inf,0);

I2 = int(P0,x,0,+inf);

alpha = int(I1,y,L1>T)+ int(I2,y,L2>T);
end

same for error2. But when I define y1 = error1(t); it says "Error in MuPAD command" I don't know what that is.

[Walter Roberson]

[Merging from a duplicate post], Maggie wrote,

I defined the following functions:

function alpha = error1(T)
syms x y
P0 = 1/(2*pi*0.6)*exp(-1/(2*0.36)*(x^2+y^2-2*0.8*x*y));
P1 = 1/(2*pi*1)*exp(-1/(2*1)*(x^2+y^2));
L1 = int(P1,x,-inf,0)/int(P0,x,-inf,0);
L2 = int(P1,x,0,+inf)/int(P0,x,0,+inf);
I1 = int(P0,x,-inf,0);
I2 = int(P0,x,0,+inf);

alpha = int(I1,y,L1>T)+ int(I2,y,L2>T);
end

function beta = error2(T)
syms x y
P0 = 1/(2*pi*0.6)*exp(-1/(2*0.36)*(x^2+y^2-2*0.8*x*y));
P1 = 1/(2*pi*1)*exp(-1/(2*1)*(x^2+y^2));
L1 = int(P1,x,-inf,0)/int(P0,x,-inf,0);
L2 = int(P1,x,0,+inf)/int(P0,x,0,+inf);
I1 = int(P1,x,-inf,0);
I2 = int(P1,x,0,+inf);

beta = int(I1,y,L1<T)+ int(I2,y,L2<T);
end

I don't if there's a problem in the expressions of alpha and beta, but what I mean is that Im trying to integrate I1 for example over the domain L1(y) < T.

After that I defined:

t = 0:0.01:30;
y1 = error1(t);

I get the following error:

Error using mupadmex
Error in MuPAD command: The argument is invalid. [Dom::Interval::new]

Error in sym/int (line 124)
   rSym = mupadmex('symobj::intdef',f.s,x.s,a.s,b.s,options);

Error in error2 (line 10)
beta = int(I1,y,L1<T)+ int(I2,y,L2<T);

Answer 1

[Walter Roberson]

You cannot provide an inequality as a bounds for int(). You can do indefinite integration by not providing a bound at all, and you can do definite integration by providing a lower bound and an upper bound, but you cannot provide an inequality and have MATLAB solve it for the domain of validity and integrate over that domain. (For one thing, defining a domain of validity does not define an integration direction, does not define which is the upper bound and which is the lower bound.)

What you need to do is to solve() the inequality first, and extract the lower bound and upper bound of range the solution is indicated as valid for, and use those bounds for your integration. See the 'ReturnConditions' option of solve() in regards to that. You might find that in practice you need to solve() as an equality rather than as an inequality, and then figure out whether the solution is the upper or lower bound and use it appropriately in integration.

Your L1 comes out as -1/(-1+erf((2/3)*y*sqrt(2))) and your L2 comes out as 1/(erf((2/3)*y*sqrt(2))+1) . If you solve(L1=T,y) then the value returned is the lower bound for L1>T. If you solve(L2==0,T) then the value returned is the lower bound for L2<T.

I notice that your forms are exactly the same for error1 and error2, with the only difference being which side of the inequality you are choosing. I recommend that you merge both of them into the same routine that returns two values, instead of having to go through all the hard work twice.

int(I2,y) has a nice closed form representation, but int(I1,y) does not, so your alpha and beta are going to be represented by expressions that include unevaluated integrals even though you may have given definite T values. That basically indicates that a numeric integral must be done at that point.

What I suggest you do is calculate error1 and error2 on a symbolic T (and merge the two into a single routine), returning symbolic formula. Then use matlabFunction() to convert the symbolic formula into an anonymous function suitable for being invoked with a definite numeric T and returning a numeric (not symbolic) result.

matlabFunction() normally produces vectorized code, so normally you would be able to invoke it on your t = 0:0.01:30 and get back a vector of results that you could plot. I am, however, not certain that matlabFunction will be able to vectorize when numeric integration is involved. I do not have the Symbolic Toolkit so I cannot test that myself. If not then the adjustments are not difficult.

Caution: the solution to the inequalities are only valid as upper or lower bounds for the case of T >= 1/2. Your L1 and L2 both have lower bounds of 1/2 for real-valued y. You need to decide whether you want to be integrating over complex y or not -- nothing in your problem statement disallows doing so, but when your boundaries are complex valued, you cannot uniquely define upper and lower bounds for the inequalities and so cannot uniquely say what you are integrating over.