Differentiating a cumulative distribution function

I am trying to numerically evaluate the integral of the normal density function of x with respect to the cumulative distribution of (x+y-m).

I am currently attempting to solve this by trying to integrate the derivative of the cdf times the pdf with respect to x.

Here is my code so far:

k=100;
m=20;
y=0;
syms x;
f=normpdf(x);
g=diff(normcdf(y-x+m);
D=simple(int(f*g,0,(k-m)))
Q=double(D);
E=normcdf(y+2*m-k)-D

Any suggestions? The error message is Error using NaN Trailing string input must be 'single' or 'double'.

Error in normcdf (line 60)
p = NaN(size(z),class(z));

Answer 1

[Tom Lane]

The Statistics Toolbox functions normpdf and normcdf don't accept symbolic inputs. You might be able to use functions like erf, which do operate on symbolic variables, to accomplish what you want to do. Here's a little example where I compute normcdf on a numeric vector, then I use Symbolic Toolbox functions to get the same result:

>> x = [.5 .7 .9];
>> normcdf(x)
ans =
    0.6915    0.7580    0.8159
>> syms s;
>> p = erfc(-s/sqrt(2))/2;
>> double(subs(p,'s',x))
ans =
    0.6915    0.7580    0.8159

[Carter]

Thanks, as I've stated below, this works, however it is through trial and error not the way I am wanting to go about this.

Answer 2

[Carter]

Thanks! You've really helped me out there. However, I've realized I need to go about this same thing differently. Learning from what you've shown,

syms x mu sigma
f=1/(sigma.*sqrt(2.*pi)).*int(exp(-(x-mu)^2)./(2.*(sigma^2)),-inf,x);
subs(f,{x mu sigma},{3.75 3.75 1}) % Should be equal to below
normcdf(3.75,3.75,1) % normcdf

I however do not get the same answer (should be .5 in this example). What did I do wrong?

Answer 3

[Tom Lane]

You just need to fix your parentheses. Change this

int(exp(-(x-mu)^2)./(2.*(sigma^2))

to this

int(exp(-(1/2)*((x-mu)/sigma)^2)