Assume $X$ is a standard Cauchy random variable with density $f(x)$, and define a continuous transformation $Z= g(X)$. What happens with

$$E(Z) = \int f(x)g(x)dx = ?$$

Does it exist /is finite? Are there some general results for classes of functions, or one has to examine the issue case-by-case, for each specific $g()$?

I understand that if for example $g(x)$ is of the form $g(x) = x^k, k \in \text{N}$, then we are looking at the raw moments of the Cauchy itself, so if $k$ is even $E(Z)$ will be infinite, while if $k$ is odd, $E(Z)$ is undefined.

Any other such conclusions?

I feel this is basic but somehow what I know (or remember) about integration theory does not seem to help me here.


2 Answers 2


I don't know of any results from integration theory that would give full description of such functions, but I can give a few observations that might be helpful.

0) If the function $g$ is always positive or always negative, then the integral exists, although it might by infinite.

1) The set of $g$ for which the integral is finite forms a linear space.

2) For any $g$ that is Lebesgue integrable the expectation exists and is finite, since: $$ |\int f(x) g(x) dx | \leq \int \frac{1}{\pi} |g(x)| dx < \infty $$

3) For any $g$ that is essentially bounded the integral is also finite since: $$ |\int f(x) g(x) dx | \leq \int |f(x)| C dx = C$$ Where $C$ is such that $g(x) \leq C$ almost everywhere.

4) If there exists $\alpha <1 $ such that $\lim_{|x| \rightarrow \infty} \frac{g(x)}{|x|^{\alpha}} = 0$ and $\int_{-n}^{n} f(x) g(x) dx$ exists for any $n > 0$, then the integral exists.

Proof : From the first assumption about $g$, there exists $n$ such that for $|x| > n$ we have $\frac{g(x)}{\pi(x^2 + 1)} \leq \frac{1}{|x|^{2 - \alpha}}$. This imiples: $$ \int f(x)g(x) dx \leq \int_{-\infty}^{-n} \frac{1}{|x|^{2 - \alpha}} dx + \int_{-n}^{n} f(x)g(x) dx + \int_{n}^{\infty} \frac{1}{|x|^{2 - \alpha}} dx < \infty$$

I suppose for most situations of interest, when the integral actually exists at least one of those criterions will work. Especially the 4th one should be quite sharp, although might not be very comfortable to check. Furthermore they are sufficient conditions for the existance of the integral, and neither of them is necessary.

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    $\begingroup$ Thanks, this is a rich answer. One question: why did you start counting cases from 0)?:) $\endgroup$ Commented Nov 1, 2014 at 20:26
  • 5
    $\begingroup$ Because the 0-th case came to my mind when I was writing the fourth one, and I didn't feel like changing the numbers :P Beside that it's so trivial it doesn't deserve a higher number :P $\endgroup$ Commented Nov 1, 2014 at 21:02

This started out as a comment (just thinking out loud here), but I got further along than I thought. I haven't read any other answers yet so there's probably some overlap.

[This is clearly not anything like a definitive answer, since it doesn't characterize when such integrals converge in general. What it does after some initial rambling is explore one set of special cases of $g$, the first couple of which may be useful in practice.]

If $g$ is small enough in the tails (and doesn't shoot off to $\infty$ too fast anywhere in the middle), the limit for the integral would be finite; $g(x)=C$, for example, is fine, as is $g(x)=\phi(x)$ (or indeed any density). On the other hand, clearly you can't have $g(x)=x$, but you could have something like $g(x) = a+bF(x)$ for some cdf.

Since $f$ is so nice near 0, as long as $g(x)$ doesn't shoot off to $\infty$ at small $x$, the fun is at the big end, so one important aspect of the question seems to be 'how fast can $g$ grow?'.

Let's split up $∫^∞_{−∞}f(x)g(x)dx$ into three parts, a left part, a right part and a middle part, and try to identify a bound for the right part only. Because $f$ is symmetric, playing with the left part is very similar to the right, and because $f$ is bounded, the behavior in the middle is only an issue if your $g$ goes off to $∞$ in that part (when you'd have to worry about how it behaved then) - but I assume you'd have mentioned if $g$ might do that.

So I assume you mostly care to bound $∫^∞_c f(x)g(x)dx$ for some sufficiently large* positive $c$.

* In the sequence of functions I consider later, I'll also need a sequence of $c$'s to go with it. In effect, each $p_j$ will have some corresponding $c_j$.

Note that if $c>1$, $\frac{1}{1+x^2}<\frac{1}{x^2}<\frac{2}{1+x^2}<\frac{2}{x^2}$, so (taking $f^*(x)=x^{-2}$), if $∫^∞_c f^*(x).g(x)dx$ diverges or converges, so does $∫^∞_c f(x)g(x)dx$, and vice-versa (they share their convergence properties). So we only need to consider this simpler case.

To simplify further, let's call $f^*(x)g(x)$ "$p(x)$" and see what $p(x)$ can do in the extreme tail ... if we get some worthwhile limits on $p(x)$ in some particular cases, we can then back out bounds on $g(x)$ from that. (Excuse me, but I'm going to be overloading the symbol "$p$" here, so there's a sequence of functions $p_j(x)$ involving $x$, $log(x)$ and a scalar power, $p$. I could call $p_j(x)$ say $q_j(x)$ and denote the dependence on the power $p$ by say $q_j(x;p)$ but to me it was more intuitive in this case to overload $p$ to show that dependence. I don't know if that's just me, though.)

By bounding the right tail of the Cauchy between multiples of $f^*(x)=\frac{1}{x^2}$, we've simplified convergence considerations to cases that are easier to examine.

For example, if we examine the behavior of $p_0(x)=1/x^p$ then we're effectively considering what happens with $g(x)=1/x^{p-2}$.

Since $I_0=\int_1^\infty 1/x^p dx$ diverges for $p\leq 1$, the required integral diverges for $g(x) = x^k$ for $k\geq1$ (a fact we already knew, of course), so to get convergence you would need $g(x)/x$ to head toward zero at some rate. We'll explore how fast it needs to do that via the mechanism of examining some functions, $p(x)$ again.

Let's now consider $I_1=\int_c^\infty 1/(x.\log(x)^p) dx$ (i.e. $p_1=1/(x\log(x)^p)$) for some finite positive $c$ that's large enough we don't have to worry about the left hand end. Some $c>1$ would do to start but we can play with it if we need to. (Here we're effectively considering a connection to $g(x)$ that on the right side is bounded by $x/\log(x)^p$, but of course we'd need some other bound for values near 0)

When $p=1$ this integral also diverges. But if $p>1$, consider the substitution $y=\log(x)$, whereupon we get:

$I_1=\int_{\log{c}}^\infty 1/(x^p) dx$

So again, it looks like this will converge if $p>1$.

We can keep iterating this trick! Consider

$I_2=\int_c^\infty 1/(x.\log(x).\log(\log(x))^p) dx$

Let $y=\log(x)$

$I_2=\int_{\log{c}}^\infty 1/(y\log(y)^p) dy$

and so again, if $p>1$ (and $c$ is not too small), that integral will converge. Similarly we could make an $I_3$, $I_4$ and so on.

So there's an interesting sequence of functions in the right tail* where you have divergence when $p\leq 1$ and convergence for $p>1$.

* $p_0(x) = \frac{1}{x^p}$, $p_1=\frac{1}{x\log(x)^p}$, $p_2=\frac{1}{x\log(x)\log(\log(x))^p}$, $\ldots$

It seems that $g(x)$ can get really close to $x$ for large $x$, since as long as $g(x)\cdot(xp_j(x))$ is bounded and $p>1$ for some $j$, $\int_c^\infty f(x)g(x)dx$ for some sufficiently large $c$ should converge to some finite bound.

In effect, we have a corresponding sequence of $g_j(x)$:

* $g_0(x) = x^{2-p}$, $g_1=\frac{x}{\log(x)^p}$, $g_2=\frac{x}{\log(x)\log(\log(x))^p}$, $g_3=\frac{x}{\log(x)\log(\log(x))\log(\log(\log(x)))^p}$, $\ldots$

for which $\int^\infty_{c_j} \frac{1}{1+x^2}g(x)dx$ will converge if $p>1$ and diverge if $p\leq 1$.

[Of course if $g$ is not bounded at finite $x$ or does something weird at the left hand end, you still have to worry about the rest of the integral.]

I've got a strong feeling I've seen a sequence like this before, so I bet this is a very well-known result.

  • $\begingroup$ That's stimulating, thanks Glen. I am a bit confused about the integrals whose convergence you examine though: what is their connection to $E(Z)$ as the latter is defined in the question? $\endgroup$ Commented Nov 1, 2014 at 22:32
  • $\begingroup$ Alecos - In effect, I'm splitting up $\int_{-\infty}^\infty f(x)g(x)dx$ into three parts, a left part, a right part and a middle part, and trying to identify a bound for the right part only. Because $f$ is symmetric, playing with the left part is very similar to the right, and because $f$ is bounded, the behavior in the middle is only an issue if your $g$ goes off to $\infty$ in that part. So I assumed you mostly cared to bound $\int_c^\infty f(x)g(x)dx$ for some positive $c$. I'll include a short section near the start with some more details explaining the connection between $p_j$ and $f.g$. $\endgroup$
    – Glen_b
    Commented Nov 1, 2014 at 22:39
  • $\begingroup$ But I can't do it immediately, unfortunately. $\endgroup$
    – Glen_b
    Commented Nov 1, 2014 at 22:43
  • $\begingroup$ Ok that, but I think I am confused by not "seeing" the Cauchy density anywhere... At your leisure. $\endgroup$ Commented Nov 1, 2014 at 22:44
  • $\begingroup$ Well, I didn't get any additional interruptions here, so I've tried to make it a bit clearer what's going on; in effect I just bound the right tail of the Cauchy above and below by $k.x^{-2}$. If there's anything I haven't explained clearly enough, please ask. I haven't tried to be formal, but I think it conveys the gist of how quickly you need $f.g$ to go down in the tail (via a set of functions that converge for $p>1$ and diverge otherwise). $\endgroup$
    – Glen_b
    Commented Nov 1, 2014 at 23:37

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