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If it is impossible, what is the proof?

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    $\begingroup$ what have you done to answer that question? $\endgroup$ – aaaaa says reinstate Monica May 27 at 21:00
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    $\begingroup$ Is this question connected with an assignment? In which case it should have come with the self-study tag. $\endgroup$ – Xi'an May 28 at 10:09
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Take two positive iid Cauchy variates $Y_1,Y_2$ with common density $$f(x)=\frac{2}{\pi}\frac{\mathbb I_{x>0}}{1+x^2}$$ and infinite expectation.

The minimum variate $\min(Y_1,Y_2)$ then has density $$g(x)=\frac{8}{\pi^2}\frac{\pi/2-\arctan(x)}{1+x^2}\mathbb I_{x>0}$$ Since (by L'Hospital's rule) $$\frac{\pi/2-\arctan(x)}{1+x^2} \equiv \frac{1}{x^3}$$ at infinity, the function $x\mapsto xg(x)$ is integrable. Hence, $\min(Y_1,Y_2)$ has a finite expectation actually equal to $\log(16)/\pi$.

More generally, in a regular Cauchy sample $X_1,\ldots,X_n$, with $n\ge 3$, every order statistic but the extremes $X_{(1)}$ and $X_{(n)}$ enjoys a (finite) expectation. (Furthermore, $X_{(1)}$ and $X_{(n)}$ both have infinite expectations, $-\infty$ and $+\infty$ resp., rather than no expectation.)

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  • $\begingroup$ Thanks. But do you mean only Xn does not enjoy finite expectation? $\endgroup$ – Preston Lui May 27 at 15:13
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    $\begingroup$ Very nice example. And $E[\min(Y_1,Y_2)] = log(16)/\pi$ $\endgroup$ – wolfies May 27 at 15:46
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Let's find a general solution for independent variables $X$ and $Y$ having CDFs $F_X$ and $F_Y,$ respectively. This will give us useful clues into what's going on, without the distraction of computing specific integrals.


Let $Z=\min(X,Y).$ Then, from basic axioms and definitions, we can work out that for any number $z,$

$$\eqalign{ F_Z(z) &= \Pr(Z\le z) = 1 - \Pr(Z > z) = 1 - \Pr(X \gt z, Y\gt z) \\&= 1 - (1-F_X(z))(1-F_Y(z)).}$$

For any CDF $F$, the expectation is

$$E_F = \int_{-\infty}^0 F(z)\mathrm{d}z + \int_{0}^\infty (1-F(z))\mathrm{d}z,$$

the sum of a negative part and a positive part.

Consequently, the question asks whether it's possible for $E_{F_Z}$ and $E_{F_Y}$ to be infinite but for $E_{F_Z}$ to be finite. This requires both the negative and positive part of $E_{F_Z}$ to be finite. Rather than analyzing this fully, it will suffice to study what happens to the positive parts: you can work out the analog for the negative parts.

In the worst case, then, the integrals $\int_0^\infty (1-F_X(z))\mathrm{d}z$ and $\int_0^\infty (1-F_Y(z))\mathrm{d}z$ will diverge but we wonder whether the integral of the product

$$\int_0^\infty (1-F_X(z))(1-F_Y(z))\mathrm{d}z$$

diverges. Clearly it cannot be any worse than the original two integrals, because since $0\le F(z)\le 1$ for all $z,$

$$\int_0^\infty (1-F_X(z))(1-F_Y(z))\mathrm{d}z \le \int_0^\infty (1-F_X(z))\mathrm{d}z \, \sup_{z\ge 0} (1-F_Y(z)) \le \int_0^\infty (1-F_X(z)).$$

This is sufficient insight to survey the landscape. Suppose that as $z\to \infty,$ $1-F_X(z)$ is approximated by $z^{-p}$ for some positive power $p,$ and similarly $1-F_Y(z)$ is approximated by $z^{-q}$ for $q \gt 0.$ We write $1-F_X \sim O(Z^p)$ and $1-F_Y \sim O(Z^q).$ Then, when both $p$ and $q$ are less than $1,$ $E_{F_X}$ and $E_{F_Y}$ are infinite.

  • When $p+q \le 1,$ because $(1-F_X)(1-F_Y)\sim O(z^{p+q}),$ $E_{F_Z}=\infty.$

  • But when $p+q \gt 1,$ $E_{F_Z}$ is finite because $\int_0^t (1-F_Z(z))\mathrm{d}z$ is bounded above by $\int_0^1 (1-F_Z(z))\mathrm{d}z$ plus some multiple of $$\int_1^t z^{-(p+q)}\mathrm{d}z = \frac{1}{p+q-1}\left(1 - t^{-(p+q-1)}\right) \to \frac{1}{p+q-1} \lt \infty.$$

In other words, the infinite expectations of the positive parts of $X$ and $Y$ imply their survival functions $1-F_X$ and $1-F_Y$ approach their lower limit of $0$ only very slowly; but the product of those survival functions, which is the survival function of $Z,$ can approach $0$ sufficiently quickly to give $Z$ a finite expectation.

In short,

For $Z$ to have finite expectation, $(1-F_X)(1-F_Y)$ must converge to $0$ sufficiently rapidly at $+\infty.$ This can happen even when neither $1-F_X$ or $1-F_Y$ converge sufficiently rapidly.

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    $\begingroup$ As an illustration we can consider the Lomax distribution with survival $S(x) = [1 + x / \beta]^{-\alpha}$ for $x > 0$ where $\beta >0$ and $\alpha >0$ are the scale and shape parameters. The expectation is finite for $\alpha > 1$. If $X$ and $Y$ are independent Lomax with the same scale $\beta >0$ and shapes $\alpha_X$, $\alpha_Y$ we see that $\min(X,\,Y)$ is Lomax with scale $\beta$ and shape $\alpha_X +\alpha_Y$. $\endgroup$ – Yves Jun 16 at 7:50
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Well, if you don't impose independance, yes.

Consider $Z \sim Cauchy$ and $B \sim Bernouilli(\frac{1}{2})$. Define $X$ and $Y$ by:

$$X = \left\{ \begin{array}[ccc] 0 0 & \text{if} & B = 0\\|Z| & \text{if} & B = 1\end{array}\right. $$

$$Y = \left\{ \begin{array}[ccc] . |Z| & \text{if} & B = 0 \\0 & \text{if} & B = 1\end{array}\right. $$

Where $|.|$ denotes absolute value. The $X$ and $Y$ have infinite expectation, but $\min(X, Y) = 0$ so $E(\min(X, Y)) = 0$.

For independent random variables, I don't know, and I would be interested in a result!

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    $\begingroup$ X and Y are not truly two dimensional case, though, I bet you can show the covariance matrix rank is 1. it's really just B you're proposing. i wouldnt count this as an example. basically X and Y are defined on the same exact event set, so it's the same variable in some sense. i's like saying I have X=Z and Y=1-Z. $\endgroup$ – Aksakal May 27 at 13:21
  • $\begingroup$ Yes... this is a bit of a cheat example ^^. I think covariance matrix is undefined on this example, but indeed the support of $(X, Y)$ is of null measure in $\mathbb{R}^2$. Having non totally linked random variables isn't hard though, with same setup, draw $X$ and $Y$ from independent uniform distribution if $B = 0 $ and from independant absolute value Cauchy if $B = 1$. $\endgroup$ – Pohoua May 27 at 13:54
  • $\begingroup$ can you show in this case that min is indeed finite? $\endgroup$ – Aksakal May 27 at 14:43
  • $\begingroup$ Well, yes, since it is always smaller than one (since there is always one of $X$ or $Y$ simulated from uniform. $\endgroup$ – Pohoua May 27 at 19:50
  • $\begingroup$ In my 2-previous comment, I made a mistake, I meant than $X\sim\mathcal{U}([0, 1])$ if $B= 0$ and $Y\sim\mathcal{U}([0, 1])$ if $B = 1$, and same for the Cauchy part of the distribution, such that for any value of $B$, one of the two variables has been generated by a $\mathcal{U}([0, 1]). $\endgroup$ – Pohoua May 27 at 21:22
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This answer is not as general as Whuber's answer, and relates to identical distributed X and Y, but I believe that it is a good addition because it gives some different intuition. The advantage of this approach is that it easily generalizes to different order statistics and to different moments or other functions $T(X)$. Also when the quantile function is known then the possibility or impossibility of 'making a statistic finite' by using an order statistic is easily seen by the type of singularity at 0 and 1.

A quick intuitive view of the possibility that an order statistic might have finite finite expectation even when the underlying variable does not can be done via the quantile function.

We can view the moments of a distribution as the moments of the quantile function: https://stats.stackexchange.com/a/365385/164061

$$E(T(x)) = \int_{0}^1 T(Q(q)) dq \\$$

Say we wish to compute the first moment then $T(x) = x$. In the image below this corresponds to the area between F and the vertical line at $x=0$ (where the area on the left side may count as negative when $T(x)<0$).

Cauchy versus Normal

The curves in the image show how much each quantile contributes in the computation. If the curve $T(Q(F))$ goes sufficiently fast enough to infinity when F approaches zero or one, then the area can be infinite.


Now, for an order statistic the integral over the quantiles $dq$ changes somewhat. For the normal variable each quantile has equal probability. For an order distribution this is beta distributed. So the integral becomes for a sample of size $n$ and using the minimum:

$$E(T(x_{(n)})) = n! \int_{0}^1 (1-q)^{n-1} T(Q(q)) dq \\$$

This term $(1-q)^{n-1}$ might be able to make a function that initially integrated to infinity because it had a pole of order 1 or higher (it's behaviour near $q=1$ was like $T(Q(q)) \sim (1-q)^{-a}$ with $a>1$), is now able to integrate to a finite value.


Example: the sample mean of the median of a sample taken from a Cauchy distributed variable is now finite because the poles of 1st order are removed. That is, $q^a(1-q)^b \tan(\pi (q-0.5))$ is finite for $a\geq 1$ and $b\geq 1$. (this relates to the more general statement of Xi'an about order statistics in relation to a Cauchy variable)

Further: When the quantile function has an essential singularity, for example $Q(p) = e^{1/(1-p)} - e$ then the sample minimum remains with infinite or undefined moments no matter the size of the sample (I just made up that quantile function as example, it relates to $f(x) = \frac{1}{(x+a)\log(x+a)^2}$, I am not sure whether there are more well known distributions that have an essential singularity in the quantile function).

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    $\begingroup$ (+1) very nice point of using the quantile transform. I would suggest writing the last integral equation as$$\mathbb E[T(X_{(n)})]=n! \int_{0}^1 (1-q)^{n-1} T(Q(q)) \text{d}q$$to make things clearer. $\endgroup$ – Xi'an May 29 at 10:18
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It's the case with almost any distribution because the expectation on a subset grows usually much slower than the subset. Let's look at the expectation on a subset for a variable $z$ with PDF $f(z)$: $$E_x[z]=\int_{-\infty}^xzf(z)dz$$ Let's look at the rate of growth of this exepctation: $$\frac d {dx}E_x[z]=xf(x)$$ So the expectation on a subset grows much slower than $x$, the boundary of a subset. The implication is that although for a distribution with no moments such as modulus of Cauchy $|z|$ the expectation is infinite $E_\infty[|z|]=\infty$, its growth with upper boundary of the subset slows down a lot with large $z$. In fact for this case $E_x[z]\approx 1/x$.

Why is this relevant? Here's why. Look at the expectation of $E[x|x<y]$ where both $x,y$ are from the same distribution with density $f(.)$ that has infinite mean: let's look at the expectation of the minimum: $$E[x|x<y]=\int_{-\infty}^\infty dyf(y)\int_{-\infty}^{y}dxf(x)\times x\\ =\int_{-\infty}^\infty dy f(y)E_y[x] $$ Since $E_y[x]$ grows much slower than $y$, this integral most likely will be finite. It is certainly finite for modulus of Cauchy $|x|$ and is equal to $\ln 4/\pi$:

  • $E_x[|z|]=\int_0^x\frac 2 \pi\frac {z}{1+z^2}dz=\int_0^x\frac 1 \pi\frac {1}{1+z^2}dz^2=\frac 1 \pi \ln(1+z^2)|_0^x=\frac{\ln(1+x^2)}{\pi}$ - here already we see how the expectation on subset slowed down from $x$ to $\ln x$.
  • $E[x|x<y]=\int_{0}^\infty \frac 2 \pi\frac 1 {1+x^2}\frac{\ln(1+x^2)}{\pi}dx =\frac 1 \pi \ln 4 $

You can apply this analysis to the minimum function trivially.

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  • $\begingroup$ Since $\mathbb E[\min(|X_1|,|X_2|)]=\log(4^2)/π$, I was puzzled by the discrepancy. I think it is because you forgot the normalising $\mathbb P(x<y)=1/2$ in all conditionings and in particular in the conditional expectations. $\endgroup$ – Xi'an May 28 at 10:15

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