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I was going through a problem in Geoffrey Grimmett and David Stirzaker's book (Probability and Random Processes). The problem is as follows:

If $f$ and $g$ are probability density functions, then prove that for $ 0 \leq \lambda \leq 1$ the function $\lambda f + (1-\lambda)g$ is a density function. Is the product $fg$ a density function as well?

It is straightforward to prove $\lambda f + (1-\lambda)g$ is a density function. For the second question as well, one can construct trivial functions for $f$ and $g$ as $f(x)=g(x)=1$ for $ 0 \leq x \leq 1$.

Are there any other non-trivial examples of a family or class of distributions for which one can find $\int_{-\infty}^{\infty} f(x)g(x) dx=1$?

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A slight extension is if $f$ has support on a set of length no greater than one and $g(x) =1$ everywhere $f$ has support. –  Jonathan Christensen Jan 23 '13 at 1:24
    
Do you mean $\int_{-\infty}^\infty f(x) g(x)\ dx = 1$ for all $f$ and $g$ in some set $S$? Cauchy-Schwarz says all members of such a set are equal a.e. –  Robert Israel Jan 23 '13 at 1:26
    
I think the second question is not asking whether there exist density functions $f$ and $g$ such that $fg$ is a density, but rather whether $fg$ is always a density function for arbitrary $f$ and $g$ the way $\lambda f + (1-\lambda)g$ is always a density function (for arbitrary choice of $\lambda \in [0,1]$). The answer is, of course, No, but there do, as you note, special cases where $fg$ is a density for particular choices of $f$ and $g$. –  Dilip Sarwate Jan 23 '13 at 1:26
    
@JonathanChristensen Well, that's slightly less trivial :). –  jay-sun Jan 23 '13 at 1:29
    
@RobertIsrael Not necessarily. Their product needs to integrate to $1$ over the real line, that's all. –  jay-sun Jan 23 '13 at 1:31

2 Answers 2

up vote 2 down vote accepted

Given any probability densities $f(x)$ and $g(x)$ with $f(x) g(x) > 0$ on a set of positive measure, and any constant $r > 0$, $r f(rx)$ and $r g(rx)$ are also probability densities, and $$\int_{\mathbb R} (r f(rx))(r g(rx))\ dx = r \int_{\mathbb R} f(x) g(x)\ dx$$ We can then choose $r$ so that this is $1$. That gives us two probability densities $r f(rx)$ and $r g(rx)$ whose product is a probability density.

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One simple construction is to let $f(x) = g(x) = k >1$ on the shared interval $[0, \frac {1}{k^2}]$, and then let $f(x), g(x)$ have disjoint support otherwise.

Note: If $f < 1 $ (or $g < 1$), then $\int fg\, dx < \int f\, dx = 1$ so we do not have a probability distribution.

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