A Integral of squared univariate PDF

[Kvad]

Hi all,

I was trying to find an answer, but couldn't, what is the integral of the squared probability density function? It doesn't seem to be equal to the square of cumulative distribution function, but how to tackle it?

∫(f(x))²dx = ? Can we transform it into, say, ∫f(x)dF(x)? and then ∫F'(x)dF(x) may be used somehow? I am not a mathematician so just throwing out some ideas I had, would appreciate any help with it :)

 

[mathman]

The transforms you have are O.K., but I can't see where you can go from there. You might try some simple pdf's and see what happens.

 

[Stephen Tashi]

I was trying to find an answer, but couldn't, what is the integral of the squared probability density function?

It isn't clear what you mean by "an answer". What requirements do you want an answer to satisfy?

The integral can be expressed in various ways.

For example, $\int (f(x))^2 dx = \int f(x) f(x) dx$, so you could use integration by parts.

Also $\int (f(x))^2 dx = \int (1)( (f(x))^2) dx$ so you could use integration by parts on that product.

 

[Kvad]

I am looking just for an analytical answer, If I do integration by parts correctly then I get this?

$F(x)f(x) = ∫f'(x)F(x)dx + ∫f(x)^2dx$

Now $∫f(x)^2dx=F(x)f(x)-∫F(x)f'(x)dx$ and I am not sure about how to tackle the last integral?

 

[Stephen Tashi]

I am looking just for an analytical answer.

What do you mean by an "analytical answer"? - an expression without integral signs? - a polynomial whose terms involve only constants and the terms $f(x)$ and $F(x)$? - a polynomial whose terms involve only constants and the terms $F(x)$, $f(x)$, and $f'(x)$ ?

If I do integration by parts correctly then I get this?

$F(x)f(x) = ∫f'(x)F(x)dx + ∫f(x)^2dx$

Yes.

Now $∫f(x)^2dx=F(x)f(x)-∫F(x)f'(x)dx$ and I am not sure about how to tackle the last integral?

I'm not sure what kind of expression you are aiming for.

 

[Kvad]

Something that will not have any integrals and derivatives, expressed in parameters and f(x) and F(x)...

 

[Stephen Tashi]

Something that will not have any integrals and derivatives, expressed in parameters and f(x) and F(x)...

I don't think that can be done (for an arbitrary f(x)) unless you use parameters that come from evaluating integrals and derivatives of f(x) at particular values of x.

For example, you can expand $g(x) = (f(x))^2$ in Taylor series about $(x-a)$ and you get a possibly infinite series of polynomial terms involving $x$ and constants, but the constants are computed by evaluating derivatives of $g(x)$ at $x = a$. If you integrate that series term-by-term you get an answer that is a series with polynomial terms.

 

[Kvad]

Well unfortunately I am interested only in the general case... But maybe if I rewrite $∫F(x)f'(x)dx$ as $∫F(x)df(x)$ getting an integration of CDF by PDF, maybe this problem is known to have some solution or on the contrary unsolvable?

I tried to substitute $f(x)=z$ and then get an easy integral $∫\frac{z^2}{2}dz$, but the solution does not match the first part if differentiated... is it wrong to do this type of substitutions?

 

[Stephen Tashi]

Well unfortunately I am interested only in the general case... But maybe if I rewrite $∫F(x)f'(x)dx$ as $∫F(x)df(x)$ getting an integration of CDF by PDF, maybe this problem is known to have some solution or on the contrary unsolvable?

I tried to substitute $f(x)=z$ and then get an easy integral $∫\frac{z^2}{2}dz$, but the solution does not match the first part if differentiated... is it wrong to do this type of substitutions?

It isn't clear what $\int F(x)df(x)$ would mean in the notation of ordinary calculus except that it denotes doing an integration by substitution. You can do the substitution $z = f(x)$ and transform the problem, but you can't simply substitute $dz$ in place of $dx$. You must use the relation $x = f^{-1}(z)$ , $dx = D_z f^{-1}(z) dz$. If $f$ is not 1-to-1, you have to break up the integration into intervals where it is 1-to-1 so that $f^{-1}$ is defined on those intervals.