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Consider a continuous random variable $F''$ following a doubly noncentral $F$ distribution with $8$ and $2$ degrees of freedom, and noncentrality parameters $0$ and $10$; which is to say:

$$ F'' \sim F_{8,2}(0,10) \text{.} $$

[Actually, $F''$ can be also seen as the reciprocal of $F'$ following a (singly) noncentral $F$ distribution with $2$ and $8$ degrees of freedom and noncentrality parameter $10$.]

The theory (and Mathematica) says that the mean of $F''$ does not exist, or it is not defined. The following chart depicts the average of $F_{8,\nu_2}(0,10)$ when $\nu_2$ varies between $0$ and $8$. It clearly shows a vertical asymptote in $\nu_2=2$, which graphically confirms that the mean of $F''$ does not exist.

IMAGE 1:

IMAGE 1

Actually, if we plot not the mean itself (which is not defined for $\nu_2 \le 2$) but any of the available analytical expressions for it or, equivalently, for the first raw moment of $F_{8,\nu_2}(0,10)$, the picture is even clearer:

IMAGE 2:

IMAGE 2

In the chart above, the following expression for the first raw moment of $F_{\nu_1,\nu_2}(\delta_1,\delta_2)$ has been used:

$$ \begin{align} \mu_1'={}& e^{-(\delta_1+\delta_2)/2} \frac{\nu_2}{\nu_1} \Gamma\!\left(\frac{\nu_1}{2}+1\right) \Gamma\!\left(\frac{\nu_2}{2}-1\right) \times \\ &{}_1\tilde{F}_1\!\left( \frac{\nu_1}{2}+1; \frac{\nu_1}{2} ; \frac{\delta_1}{2}\right){}_1\tilde{F}_1\!\left( \frac{\nu_2}{2}-1; \frac{\nu_2}{2} ; \frac{\delta_2}{2}\right) \text{,} \end{align} $$

as seen in http://mathworld.wolfram.com/NoncentralF-Distribution.html.

However, when performing simulations on the mean of $F''$, things seem —I am not sure— to converge to a reasonable value between $0.2$ and $0.4$:

IMAGE 3:

IMAGE 3

The chart above is the result of calculating the sample average of up to $100\,000$ random observations of $F'' \sim F_{8,2}(0,10)$. The sample average of the $100\,000$ observations happens to be $0.338527$.

Although the theoretical mean of the distribution does not exist, is there a way to give a suitable substitute for it (let us call it pseudomean, for instance)? Clearly, I am not asking for the median, since the distribution is not symmetrical.

For instance, is there a way to calculate a theoretical value for the limit of these sample averages when the sample size tends to infinity (in case such a limit exists)?

To clarify, I want to add another example with $F'' \sim F_{48,\nu_2}(0,50)$:

IMAGE 4:

IMAGE 4

Despite that, in the above image, there seems not to be any discontinuity, it is there (zoom in):

IMAGE 5:

IMAGE 5

So, in this example, a suitable substitute for the nonexistent theoretical mean of the distribution $F'' \sim F_{48,2}(0,50)$ would be something between $0.041745$ and $0.041748$, more or less.

Actually, the average of a $200\,000$-element random sample of this distribution was found to be $0.0417525$. This is the associated chart:

IMAGE 6:

IMAGE 6

So, any suggestions?? My theoretical background tells me that it makes no sense trying to estimate what it does not exist, but simulations and charts suggest that there may be a reasonable theoretical substitute for the unavailable mean.

NOTE: The value should be theoretical in the sense that it should not depend on any sampling process, but only on the parameters of the distribution.


UPDATE

Actually, some related questions are:

  • What is the average of $\frac{1}{k}(X_1 + \cdots + X_k)$ when $X_i \sim F_{\nu_1,2}(0,\delta_2)$ for all $i$?

  • What is the limit of the above expression when $k$ tends to infinity?

  • Can the sample average have an average value when the initial distribution does not have one?


UPDATE

Answering the comment by @Glen_b below, a reasonable substitute for the nonexistent mean of $F_{\nu_1,2}(0, \delta_2)$ would be something that fitted the gap that is clearly observed in images 4 and 5; something between $0.041745$ and $0.041748$, in that concrete example, as I already said.

What am I going to use this pseudomean for? I am going to use it as a reference value to calculate a sort of probability intervals (something like $[ \mathrm{mean}-\mathrm{something} \, , \, \mathrm{mean}+\mathrm{something_{~}else}]$). Taking the mean as the reference point for building that interval is a convention that I would like to keep for analogy with other related cases in which the true mean does exist.

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    $\begingroup$ Since you do not want a madian, maybe a trimmed mean. Maybe you can find (possibly unequal) trimming fractions $\alpha_1,\alpha_2$ so that there is about equal "contribution to mean" trimmed from each tail? Another possibility is the first L-moment (search this site!) $\endgroup$ – kjetil b halvorsen Nov 21 '17 at 15:47
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    $\begingroup$ These questions have been answered elsewhere, typically using a Cauchy distribution as the epitome of one with an undefined mean. See stats.stackexchange.com/… for some of those posts. An undefined expectation implies all three of your questions are answered in the negative: the average has no expectation; it therefore cannot converge to some limit; and it cannot have an "average value" because that's its expectation. In short, everything follows immediately from the fact that the expectation of an average is the average of the expectations. $\endgroup$ – whuber Nov 21 '17 at 16:13
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    $\begingroup$ Thank you for both coments (@kjetilbhalvorsen and @whuber). You suggested me to try with the Trimmed Mean or L-moments. What about the Wisorised Mean? Do you think it is useful? In the case of trimmed or winsorised means, how to decide the fractions of probability to remove or the cutoffs? $\endgroup$ – Vicent Nov 21 '17 at 19:45
  • $\begingroup$ Winsorized means could be another possibility. $\endgroup$ – kjetil b halvorsen Nov 21 '17 at 21:23
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    $\begingroup$ It would be impossible to suggest a suitable substitute without some idea of the properties you want it to have. What are you using this substitute for the mean to do? $\endgroup$ – Glen_b Nov 22 '17 at 1:53

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