find finverse of cumtrapz()

Hi,

I have to invert the function that calculates cumtrapz of an equation as follows:

    mu = 0;
    var = 1;
    %x = -9:0.01:9;
    syms x
    f(x)  = cumtrapz((2*pi*(v))^(-0.5)* exp(-((x-mu).^2)/(2*(v))))
    i = finverse(f)
    fplot(i)
    

But this gives me :

f(x) =
0
Warning: Unable to find functional inverse. 
> In symengine
  In sym/privBinaryOp (line 1032)
  In sym/finverse (line 40)
  In inverse_cdf (line 7) 
i(x) =
Empty sym: 0-by-1

However the cumtrapz function does work:

    mu = 0;
    var = 1;
    x = -9:0.01:9;
    y = (2*pi*(v))^(-0.5)* exp(-((x-mu).^2)/(2*(v)));
    i = cumtrapz(y);
    plot(i)  
    

what can i do to invert this function

Answer 1

[Star Strider]

Since ‘f’:

mu = 0;
v = 1;
syms x
f(x) = int((2*pi*(v))^(-0.5)* exp(-((x-mu).^2)/(2*(v))))

evaluates to:

f(x) =
(7186705221432913*2^(1/2)*pi^(1/2)*erf((2^(1/2)*x)/2))/36028797018963968

you will need to use the erfinv function to calculate the inverse.

[Aishwarya Radhakrishnan]

I'm actually trying to get the inverse of cumulative pdf, but it is giving me proper cdf using cumtrapz but when i declare it as a function and try to find inverse, im getting error function, how can i find the inverse of cdf for univariate pdf?

[Aishwarya Radhakrishnan]

the cumtrapz calculates the cumulative pdf, but with the inverse function of the cdf, i want to get x in terms of y,

eg, y = F(x)

then x = Finverse(y)

like in figure:

but on using finverse(), im getting error as output

[Star Strider]

If you want to use cumtrapz, the interp1 function is likely the best option:

mu = 0;
v = 1;
x = -7:0.01:7;
fx = cumtrapz(1/sqrt(2*pi*(v)) * exp(-((x-mu).^2)/(2*(v))));
y = [5 25 50 75 95];
fi = interp1(fx, x, y, 'linear','extrap')

figure
plot(x, fx)
hold on
plot(fi, y, '+')
hold off
grid

This plots ‘+’ markers at the appropriate values of the ‘y’ vector. I call the inverse ‘fi’. I had to restrict your original ‘x’ vector because with the original vector, the ‘fx’ points were not unique, as interp1 defines that.