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It is well known that given a real-valued random variable $X$ with pdf $f$, the mean of $X$ (if it exists) is found by \begin{equation} \mathbb{E}[X]=\int_{\mathbb{R}}x\,f(x)\,\mathrm{d}x\,. \end{equation}

General question: Now, if one cannot solve the above integral in closed form but wants to simply determine if the mean exists and is finite, is there a way to prove that? Is there (perhaps) some test I can apply to the integrand to determine if certain criteria are met for the mean to exist?

Application specific question: I have the following pdf for which I want to determine if the mean exists: \begin{equation} f(x)=\frac{|\sigma_{2}^{2}\mu_{1}x+\mu_{2}\sigma_{1}^{2}|}{\sigma_{1}^{3}\sigma_{2}^{3}a^{3}(x)}\,\phi\left(\frac{\mu_{2}x-\mu_{1}}{\sigma_{1}\sigma_{2}a(x)}\right)\qquad \text{for}\ x\in\mathbb{R}\,, \end{equation}

where $\mu_{1},\mu_{2}\in\mathbb{R}$, $\sigma_{1},\sigma_{2}>0$, $a(x)=\left(\frac{x^{2}}{\sigma_{1}^{2}}+\frac{1}{\sigma_{2}^{2}}\right)^{1/2}$, and $\phi(g(x))=\frac{1}{\sqrt{2\pi}}\,e^{-g^{2}(x)/2}$.

I have tried to solve for the mean to no avail.

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    $\begingroup$ in your specific question $f(x)$ is not a proper density function. suppose $\mu_1 =1$, $\mu_2=0$ and $\sigma_j = 1$, $j=1,2$, then $f(x)<0$ for $x<0$. $\endgroup$ – EliKa Jul 7 '17 at 12:46
  • $\begingroup$ @EliKa Good find. There may be a typo. I will check and correct the question. That said, I am still mostly interested in the "how" part of the question, i.e. how would I got about determining if the mean exists and is finite? $\endgroup$ – Aaron Hendrickson Jul 7 '17 at 13:11
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    $\begingroup$ You could try bounding $\lvert x f(x) \rvert$ above and below by some nonnegative functions $u(x)$ and $b(x)$ such that you can integrate them. If you can integrate $u(x)$, then your distribution has a mean. If $\int b(x)dx = \infty$, then your distribution has no mean. $\endgroup$ – Ceph Jul 7 '17 at 13:33
  • $\begingroup$ @Ceph That's a good suggestion. Is that technique based on the "squeeze theorem"? $\endgroup$ – Aaron Hendrickson Jul 7 '17 at 13:36
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    $\begingroup$ @AaronHendrickson Similar idea, but (as I understand it) the squeeze theorem is a little different. Using the ST here might look like this: you find $u(x)$ and $b(x)$ that bound $xf(x)$ (rather than bounding $\lvert x f(x) \rvert$ as in my earlier comment) such that you can find $\int u(x) dx = \int b(x) dx= \mu$, where $\mu $ is the mean of your distribution. But that is probably not a plausible strategy, since you would be hard pressed to find such $u$ and $b$. (They could differ from $xf(x)$ only on a set of measure 0 and so would probably not be any easier to integrate than $xf(x)$ is.) $\endgroup$ – Ceph Jul 7 '17 at 13:42
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There is no general technique, but there are some simple principles. One is to study the tail behavior of $f$ by comparing it to tractable functions.

By definition, the expectation is the double limit (as $y$ and $z$ vary independently)

$$E_{y,z}[f] = \lim_{y\to-\infty,z\to\infty}\int_y^z x f(x) dx = \lim_{y\to-\infty}\int_y^0 x f(x) dx+ \lim_{z\to\infty}\int_0^z x f(x) dx.$$

The treatment of the two integrals at the right is the same, so let's focus on the positive one. One behavior of $f$ that assures a limiting value is to compare it to the power $x^{-p}$. Suppose $p$ is a number for which $$\liminf_{x\to\infty} x^p f(x)\gt 0.$$ This means there exists an $\epsilon\gt 0$ and an $N\gt 1$ for which $x^p f(x) \ge \epsilon$ whenever $x\in[N,\infty)$. We may exploit this inequality by breaking the integration into the regions where $x\lt N$ and $x \ge N$ and applying it in the second region:

$$\eqalign{ \int_0^z x f(x) dx &=\int_0^{N} x f(x) dx + \int_{N}^z x f(x) dx \\ &=\int_0^{N} x f(x) dx + \int_{N}^z x^{1-p} \left(x^p f(x)\right) dx \\ &\ge \int_0^{N} x f(x) dx + \int_{N}^z x^{1-p} \left(\epsilon\right) dx \\ &= \int_0^{N} x f(x) dx + \frac{\epsilon}{2-p}\left(z^{2-p} - {N}^{2-p}\right). }$$

Provided $p\lt 2$, the right hand side diverges as $z\to\infty$. When $p=2$ the integral evaluates to the logarithm,

$$\int_{N}^z x^{1-2} \left(\epsilon\right) dx = \epsilon \left(\log(z) - \log(N)\right),$$

which also diverges.

Comparable analysis shows that if $|x|^pf(x)\to 0$ for $p\gt 2$, then $E[X]$ exists. Similarly we may test whether any moment of $X$ exists: for $\alpha\gt 0$, the expectation of $|X|^\alpha$ exists when $|x|^{p+\alpha}f(x)\to 0$ for some $p\gt 1$ and does not exist when $\liminf |x|^{p+\alpha}f(x)\gt 0$ for some $p \le 1$. This addresses the "general question."

Let's apply this insight to the question. By inspection it is clear that $a(x)\approx |x|/\sigma_1$ for large $|x|$. In evaluating $f$, we may therefore drop any additive terms that will eventually be swamped by $|x|$. Thus, up to a nonzero constant, for $x\gt 0$

$$f(x) \approx \frac{\mu_1 x}{\sigma_2 x^3}\phi\left(\frac{\mu_2 x}{\sigma_2 x}\right) = x^{-2}\frac{\mu_1}{\sigma_2}\exp\left(\left(-\frac{\mu_2}{2\sigma_2}\right)^2\right).$$

Thus $x^2 f(x)$ approaches a nonzero constant. By the preceding result, the expectation diverges.

Since $2$ is the smallest value of $p$ that works in this argument--$|x|^pf(x)$ will go to zero as $|x|\to\infty$ for any $p\lt 2$--it is clear (and a more detailed analysis of $f$ will confirm) that the rate of divergence is logarithmic. That is, for large $|y|$ and $|z|$, $E_{y,z}[f]$ can be closely approximated by a linear combination of $\log(|y|)$ and $\log(|z|)$.

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