normpdf is taking zeros

Hello, I am using the normpdf command to evaluate a function in MATLAB. When the argument of normpdf is too small, MATLAB reports a zero. Is is possible to get the exact value of normpdf instead of 0?

[David Hill]

normpdf(-38);%this reports 1.097221051994971e-314, you want smaller than this?

[hey yo]

Yes, for example normpdf(-77) returns 0. It is important for me to get the exact number because I multiply this value with other numbers to find the solution to a complicated function. Thank you.

[Anjali]

Hey, I'm stuck on similar problem and not able to decode it, how did you proceed further

[John D'Errico]

@Anjali

I showed what you can do in my answer. If you will form the product of numbers, AND some of those numbers in the product are hugely large, you can always take the logs of each term in the product, then add the logs. Then exponentiate the sum at the end.

And you should see that the log of a normal PDF is trivially easy to compute. For example, even without using syms, we can do that:

mu = 0;
sig = 1;
x = -77;
logp = -(x-mu).^2/sig^2/2 - log(sig) - log(sqrt(2*pi))

logp = -2.9654e+03

Of course, that is a pretty large negative number. But now you can work with the logs of those terms. In the end, the result will still probably be numerical garbage, and still very likely result in an underflow, becaluse logp is such a negative result. But unless you are willing to work entirely in terms of higher precision numbers, this is the best you can do.

Answer 1

[John D'Errico]

No. At least, not in double precision. The "exact" value there is impossible to represent in a double precision number.

The normal pdf at z = -77 should be:

mu = 0;
sig = 1;
x = sym(-77);
p = exp(-(x-mu).^2/sig^2/2)*1/(sig*sqrt(2*sym('pi')))

p = 

vpa(p)

ans = 

Do you understand this is a number with around 1288 zeros after the decimal point, before you see any digits?

As such, the result underflows in double precision. You get ZERO.

realmin

ans = 2.2251e-308

Actually, you can go about as far as -38 or so, before an underflow occurs, with the result as what is known as a denormalized number. But that is around the limit. And even most computations with those numbers at that level will still yield numerical garbage.

normpdf(-38)

ans = 1.0972e-314

You can want what you want, but this is not possible working in double precision. If you are willing to use higher precision tools, such as syms, then yes, you can.

[hey yo]

Thank you for taking the time to post this reply. It is very valuable.