Jensen-variance distance measure: a unified framework for statistical and information measures

Abstract

The Jensen-variance (JV) distance measure is introduced and some properties are developed. The JV distance measure can be expressed using two interesting representations: the first one is based on mixture covariances, and the second one is in terms of the scaled variance of the absolute difference of two random variables. The connections between the JV distance measure and some well-known information measures, such as Fisher information, Gini mean difference, cumulative residual entropy, Fano factor, varentropy, varextropy, and chi-square distance measures, are examined. Specifically, the JV distance measure possesses metric properties and unifies most of the information measures within a general framework. It also includes variance and conditional variance as special cases. Furthermore, an extension of the JV distance measure in terms of transformed variables is provided. Finally, to demonstrate the usefulness of proposed methods, JV distance is applied to a real-life dataset related to fish condition factor index and some numerical results assuming skew-normal-distributed samples are presented.

Appendices

Appendix A: $(\alpha ,w)$-Jensen-variance distance measure

Definition 4

The $(\alpha ,w)-$JV distance between two random variables X and Y, for $\alpha$ and w $\in (0,1)$, is defined as

$$\begin{aligned} \mathcal{J}_{w,\alpha }(X,Y)= & {} w Var\big ((1-\alpha )X+\alpha Y\big )+(1-w)Var\big (\alpha X+(1-\alpha )Y\big )\nonumber \\{} & {} -Var\big ((1-\bar{s})X+\bar{s}Y\big ), \end{aligned}$$

(18)

where ${\bar{s}}=w\alpha +(1-w)(1-\alpha )$.

Theorem 10

The connection between $(\alpha ,w)-$JV distance and JV one is given by

$$\begin{aligned} \mathcal{J}_{w,\alpha }(X,Y)=(1-2\alpha )^2 \mathcal{J}_w(X,Y). \end{aligned}$$

(19)

Proof

From the definition of $\mathcal{J}_{w,\alpha }(X,Y)$ measure in (18) and upon making use Theorem 3, we have

$$\begin{aligned} \mathcal{J}_{w,\alpha }(X,Y)= & {} w(1-w)Var\big ( (1-\alpha )X+\alpha Y-(\alpha X+(1-\alpha )Y) \big )\\= & {} w(1-w) Var\big ( (1-2\alpha )(X-Y)\big )\\= & {} (1-2\alpha )^2 w(1-w)Var(X-Y)\\= & {} (1-2\alpha )^2\mathcal{J}_w(X,Y), \end{aligned}$$

as required. $\square$

Corollary 4

If X and Y be two independent random variables, then from the inequality in (5) and Theorem 10, we find that

$$\begin{aligned} \mathcal{J}_{w,\alpha }(X,Y)\le & {} \mathcal{J}_w(X,Y)\\\le & {} \frac{Var(X)+Var(Y)}{4}. \end{aligned}$$

Appendix B: Connection between Jensen-variance distance and Fano factor measure

Let X be a random variable with finite mean E[X] and variance Var(X). Then, the ratio

$$\begin{aligned} \mathcal{F}(X)=\frac{Var(X)}{E[X]} \end{aligned}$$

is known as Fano factor (Feller and Morse 1958; Cover and Thomas 2006). In fact, the Fano factor is a normalized measure used to quantify the dispersion of event occurrences within a given time window. It is often used to analyze and characterize the variability or clustering behavior of events, particularly in scenarios where events occur randomly or intermittently.

Lemma 2

Then connection between JV distance and Fano factor measure is given by

$$\begin{aligned} \mathcal{F}(X)=\mathcal{J}_\alpha \bigg (\frac{X}{\sqrt{\alpha (1-\alpha )E[X]}},W\bigg ), \end{aligned}$$

where W is a degenerate random variable in point $E[W]=c$.

Proof

From the definition of JV distance and making use Theorem 3, we have

$$\begin{aligned} \mathcal{J}_\alpha \bigg (\frac{X}{\sqrt{\alpha (1-\alpha )E[X]}},W\bigg )= & {} \alpha (1-\alpha )Var\bigg (\frac{X}{\sqrt{\alpha (1-\alpha )E[X]}}-W\bigg )\\= & {} \frac{Var(X-W)}{E[X]}\\= & {} \frac{Var(X)}{E[X]}\\= & {} \mathcal{F}(X), \end{aligned}$$

as required. $\square$

Theorem 11

Let X and Y be two random variables. Then, a connection between JV measure and Fano factor is given by

$$\begin{aligned} \frac{\mathcal{J}_\alpha (X,Y)}{\mu _\alpha }=\Lambda \mathcal{F}(X)+(1-\Lambda )\mathcal{F}(Y)-\mathcal{F}\big (\alpha X+(1-\alpha )Y\big ), \end{aligned}$$

(20)

where $\Lambda =\frac{\alpha E[X]}{\mu _\alpha }$ and $\mu _\alpha =\alpha E[X]+(1-\alpha )E[Y].$

Proof

From the definition of ${\mathcal{J}_\alpha (X,Y)}$ and considering $\Lambda =\frac{\alpha E[X]}{\mu _\alpha }$ and $\mu _\alpha =\alpha E[X]+(1-\alpha )E[Y]$, we have

$$\begin{aligned} \mathcal{J}_\alpha (X,Y)= & {} \alpha Var(X)+(1-\alpha )Var(Y)-Var\big (\alpha X+(1-\alpha )Y\big )\\= & {} \alpha E[X]\frac{Var(X)}{ E[X]}+(1-\alpha ) E[Y] \frac{Var(Y)}{E[Y]}-\mu _\alpha \frac{Var\big (\alpha X+(1-\alpha )Y\big )}{\mu _\alpha }\\= & {} \mu _\alpha \bigg \{\Lambda \frac{Var(X)}{ E[X]}+(1-\Lambda )\frac{Var(Y)}{E[Y]}- \frac{Var\big (\alpha X+(1-\alpha )Y\big )}{\mu _\alpha }\bigg \}\\= & {} \mu _\alpha \bigg \{\Lambda \mathcal{F}(X)+(1-\Lambda )\mathcal{F}(Y)-\mathcal{F}\big (\alpha X+(1-\alpha )Y\big )\bigg \}, \end{aligned}$$

as required. $\square$

From Theorem 11, it is clear that

$$\begin{aligned} \mathcal{F}\big (\alpha X+(1-\alpha )Y\big )\le \Lambda \mathcal{F}(X)+(1-\Lambda )\mathcal{F}(Y). \end{aligned}$$