Find inverse of a cdf so as to sample x

I have a cdf calculated as follows:

    mu = 0;
    var = 1;
    x = -10:0.01:10;
    y = cumtrapz((2*pi*(var))^(-0.5)* exp(-((x-mu).^2)/(2*(var))));

This gives the cdf of my probability distribution function (pdf): (2*pi*(var))^(-0.5)* exp(-((x-mu).^2)/(2*(var)))

But i want to sample the values of x randomly for this pdf. So I've created the cdf stored in variable y.

If i can find the inverse cdf, then i can uniformly select y values and then calculate using inverse cdf the corresponding x values that become my sample.

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:

I want to pass in y in the figure to Finverse(y) which is inverse of cdf, and then i can get x as:

x = Finverse(y)

But, i've tried many methods till now, but none seem to help.

i've tried to use :

    i = erfinv(y)

but this gives me values as vector, but i need function so that i can pass in the values of y to get x. Please help.

Answer 1

[Star Strider]

As I suggested earlier in find finverse of cumtrapz():

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.

Enlarging on that:

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];
Finverse = @(y) interp1(fx, x, y, 'linear','extrap');

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

That should do what you want.

[Aishwarya Radhakrishnan]

Ohh thank you so much !

[Star Strider]

As always, my pleasure!