Normal distribution and unit

I think the unit of PDF of normal distribution should be unitless or dimensionless.

http://upload.wikimedia.org/math/d/6...9d037a126c.png

But, I found a sigma in the denominator of PDF of normal distribution. So, the sigma should be unitless.

But, the sigma is the standard deviation of the random variable. This means the sigma and the random variable are also unitless.

Is this correct?

If so, can we apply normal distribution to any random variable with unit, e.g. meter or kg?

If we can do this, the unit of PDF of normal distribution would be 1/meter or 1/kg due to the existence of sigma in the equation.

I think this is not a correct expression of unit of probability.

Could any one help explain?(Nod)

Re: Normal distribution and unit

Hey eulerian.

What you are dealing with functions of any kind, things are chosen to be dimension-less if you are dealing with transcendental functions (like exponentials, trig functions, and others).

The distribution itself is going to represent a probability which is always unit-less (it is a number). If you are talking about some probability with respect to another unit, then that is a different quantity in itself.

For more information, you should probably read an applied math book, engineering book, or physics book on units and they come into play with functions, transformations, and relations to other units.

Re: Normal distribution and unit

Quote:

Originally Posted by

**eulerian** I think the unit of PDF of normal distribution should be unitless or dimensionless.

http://upload.wikimedia.org/math/d/6...9d037a126c.png
But, I found a sigma in the denominator of PDF of normal distribution. So, the sigma should be unitless.

But, the sigma is the standard deviation of the random variable. This means the sigma and the random variable are also unitless.

Is this correct?

If so, can we apply normal distribution to any random variable with unit, e.g. meter or kg?

If we can do this, the unit of PDF of normal distribution would be 1/meter or 1/kg due to the existence of sigma in the equation.

I think this is not a correct expression of unit of probability.

Could any one help explain?(Nod)

let p(x) be any probability distribution. p(x) dx represents an infinitesimal probability and thus is unitless. But dx represents a infinitesimal bit of a random variable that can have units. Thus p(x) must have units 1/(units of x).

For the Normal distribution you can see this from the standard deviation in the denominator. The std has units of the underlying random variable. The exponential term is unitless. So we end up with 1/(units of std) = 1/(units of underlying rv) as the units of p(x).

Re: Normal distribution and unit

Quote:

Originally Posted by

**romsek** let p(x) be any probability distribution. p(x) dx represents an infinitesimal probability and thus is unitless. But dx represents a infinitesimal bit of a random variable that can have units. Thus p(x) must have units 1/(units of x).

For the Normal distribution you can see this from the standard deviation in the denominator. The std has units of the underlying random variable. The exponential term is unitless. So we end up with 1/(units of std) = 1/(units of underlying rv) as the units of p(x).

Thank you.

So, the unit of likelihood function should also be 1/(units of underlying rv).

As I know, log function is a transcendental function, which only deals with unitless variables.

What is the unit of log-likelihood function? Can we log 1/(units of underlying rv)?

Re: Normal distribution and unit

Quote:

Originally Posted by

**chiro** Hey eulerian.

What you are dealing with functions of any kind, things are chosen to be dimension-less if you are dealing with transcendental functions (like exponentials, trig functions, and others).

The distribution itself is going to represent a probability which is always unit-less (it is a number). If you are talking about some probability with respect to another unit, then that is a different quantity in itself.

For more information, you should probably read an applied math book, engineering book, or physics book on units and they come into play with functions, transformations, and relations to other units.

Do you agree with #3 and #4?

Thank you

Re: Normal distribution and unit

Quote:

Originally Posted by

**eulerian** Thank you.

So, the unit of likelihood function should be 1/(units of underlying rv).

As I know, log function is a transcendental function, which only deals with unitless variables.

What is the unit of log-likelihood function? Can we log 1/(units of underlying rv)?

If I recall the likelihood function is the ratio of two of the same type of distribution and thus will be unitless as it must be to enable the log likelihood function to exist.

The log likelihood function will be unitless as well.

Re: Normal distribution and unit

Quote:

Originally Posted by

**romsek** If I recall the likelihood function is the ratio of two of the same type of distribution and thus will be unitless as it must be to enable the log likelihood function to exist.

The log likelihood function will be unitless as well.

Thank you so much.

So, in maximum-likelihood estimation, the sigma must be unitless and be a ratio?

http://upload.wikimedia.org/math/4/8...8f7eb26b76.png

Re: Normal distribution and unit

Quote:

Originally Posted by

**eulerian**

it's been a long time since I looked at something like that. I'm going to have to review it a bit.

Re: Normal distribution and unit

Any opinion is appreciated

Re: Normal distribution and unit

It looks like they are just taking the log of the standard deviation w/o regard to it's units. I'd have to read up on this a bit to give you a justification. This formula is for finding the maximum likelihood estimates of the mean and variance given a set of sample data?

I don't think I'd worry too much about the units of the likelihood function. It's not a physical thing.

Re: Normal distribution and unit

Re: Normal distribution and unit

Quote:

Originally Posted by

**eulerian**

Ok, I see. The log of the distribution is basically the entropy of the underlying random variable. I guess it's units would be in nats.

I'm not seeing any discussions of the units of the log-likelihood function jumping out at me off the web.

Re: Normal distribution and unit

Also, if 1/(units of underlying rv) is the units of p(x), how can we get the below?

http://upload.wikimedia.org/math/9/f...4dfaedc20e.png

Because Σ(p(x)x)=E[X], the unit should be $, not unitless.

Thank you.

Re: Normal distribution and unit

Quote:

Originally Posted by

**eulerian**

For continuous rvs you have E[X] = Integral[x p(x) dx, -inf, inf] and that dx contributes units.

for discrete rvs the p_{k}'s are probabilities, not probability densities and thus are unitless. Thus Sum[k p_{k}, 0, inf] has units of k, in this case dollars

Re: Normal distribution and unit

Quote:

Originally Posted by

**romsek** For continuous rvs you have E[X] = Integral[x p(x) dx, -inf, inf] and that dx contributes units.

for discrete rvs the p_{k}'s are probabilities, not probability densities and thus are unitless. Thus Sum[k p_{k}, 0, inf] has units of k, in this case dollars

Thank you so much again.

I get it.

But, I have not learned information entropy. Could you just show out some dimensional analysis about the operation between any other units and the units of information(e.g. bits or nats)?