Open Access

On a probability inequality of van Dam

Journal of Inequalities and Applications20152015:128

https://doi.org/10.1186/s13660-015-0652-1

Received: 3 July 2014

Accepted: 31 March 2015

Published: 11 April 2015

Abstract

This note derives an interesting probability inequality between the expectation of a conditional variance and the variance of a conditional expectation for a function of two independent random variables. In the special case of finite discrete random variables, the inequality coincides with an inequality by Feng and Tonge (2000), which extends a result by van Dam (1998).

Keywords

probability inequalityrandom variablevan Dam’s inequality

MSC

26E60
Let T be a continuous random variable having the probability density function ϕ defined on R. By definition
$$ E(T)=\int_{-\infty}^{\infty} t\phi(t)\,dt $$
(1)
is the expectation of T, and
$$ \sigma^{2}(T)=\int_{-\infty}^{\infty} \bigl(t-E(T)\bigr)^{2}\phi(t)\,dt $$
(2)
is the variance \(\sigma^{2}(T)\).

Let X, Y be two independent random variables with known distribution having the probability density function \(\phi_{1}(x)\) and \(\phi_{2}(y)\), respectively. Then the joint probability density function of X and Y is \(\phi_{1}(x) \phi_{2}(y)\).

Let another random variable Z be a function of X and Y
$$Z=f(X,Y), $$
where \(f(\cdot,\cdot)\in L^{2}(R^{2})\). Then the expectation of Z is
$$E(Z)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y) \phi_{1}(x)\phi _{2}(y) \,dx\,dy, $$
and the conditional probability density functions of Z, given the occurrence of the value x of X and y of Y, are equal to \(\phi _{2}(y)\) and \(\phi_{1}(x)\), respectively, such that the conditional expectations of Z are equal to
$$E(Z|X)=\int_{-\infty}^{\infty} f(x,y) \phi_{2}(y)\,dy, \qquad E(Z|Y)=\int_{-\infty}^{\infty} f(x,y) \phi_{1}(x)\,dx. $$
Furthermore, we have
$$\begin{aligned}& \begin{aligned}[b] E\bigl(E(Z|X)\bigr) &= \int_{-\infty}^{\infty} \biggl(\int _{-\infty}^{\infty} f(x,y) \phi_{2}(y)\,dy \biggr) \phi_{1}(x)\,dx\\ & = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) \phi_{2}(y)\phi_{1}(x)\,dx \,dy =E(Z), \end{aligned}\\& \begin{aligned}[b] E\bigl(Z \cdot E(Z|X)\bigr) &= \int_{-\infty}^{\infty}\int _{-\infty}^{\infty} f(x,y) \biggl(\int_{-\infty}^{\infty} f(x,t) \phi _{2}(t)\,dt \biggr) \phi_{1}(x) \phi_{2}(y) \,dx\,dy\\ & = \int_{-\infty}^{\infty} \biggl(\int _{-\infty}^{\infty} f(x,y) \phi_{2}(y)\,dy \biggr)^{2}\phi_{1}(x)\,dx =E\bigl(\bigl[E(Z|X) \bigr]^{2}\bigr), \end{aligned} \end{aligned}$$
and
$$\begin{aligned} &E\bigl(E(Z|X)\cdot E(Z|Y)\bigr) \\ &\quad= \int_{-\infty}^{\infty }\int_{-\infty}^{\infty} \biggl(\int_{-\infty}^{\infty} f(s,y) \phi_{1}(s)\,ds \biggr) \biggl(\int_{-\infty}^{\infty} f(x,t) \phi_{2}(t)\,dt \biggr) \phi_{1}(x)\phi _{2}(y) \,dx\,dy \\ &\quad = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(s,y) \phi_{1}(s) \phi_{2}(y)\,ds \,dy \int_{-\infty}^{\infty} \int_{-\infty }^{\infty} f(x,t) \phi_{1}(x) \phi_{2}(t)\,dx \,dt =\bigl[E(Z)\bigr]^{2}. \end{aligned}$$
Shortly,
$$\begin{aligned}& E\bigl(E(Z|X)\bigr)= E(Z), \end{aligned}$$
(3)
$$\begin{aligned}& E\bigl(Z \cdot E(Z|X)\bigr) = E\bigl(\bigl[E(Z|X ) \bigr]^{2}\bigr), \end{aligned}$$
(4)
$$\begin{aligned}& E\bigl(E(Z|X)\cdot E(Z|Y)\bigr) = \bigl[E(Z) \bigr]^{2}. \end{aligned}$$
(5)

Theorem 1

Let X, Y be two independent random variables and \(Z=f(X,Y)\) be a function of the random variables X and Y. Then
$$ E\bigl(\sigma^{2}(Z|X)\bigr)\leq\sigma^{2} \bigl(E(Z|Y)\bigr). $$
(6)

Proof

For convenience, we set \(Z_{1}= E(Z|X) \) and \(Z_{2} = E(Z|Y)\). It follows from (3) that
$$E(Z-Z_{1}-Z_{2})=-E(Z). $$
Since \(\sigma^{2}(Z-Z_{1}-Z_{2})=E([Z-Z_{1}-Z_{2}]^{2})-[E(Z-Z_{1}-Z_{2})]^{2}\geq0\),
$$ \bigl[E(Z)\bigr]^{2}\leq E\bigl([Z-Z_{1}-Z_{2}]^{2} \bigr). $$
(7)
From (4) and (5), we have
$$E(ZZ_{1}) =E\bigl(Z_{1}^{2}\bigr),\qquad E(ZZ_{2})=E\bigl(Z_{2}^{2}\bigr) $$
and
$$E(Z_{1}Z_{2}) = \bigl[E(Z)\bigr]^{2}. $$
Therefore, we have
$$\begin{aligned} & E \bigl([Z-Z_{1}-Z_{2}]^{2} \bigr) \\ &\quad= E \bigl( Z^{2}+Z_{1}^{2}+Z_{2}^{2}-2ZZ_{1}-2ZZ_{2}+2Z_{1}Z_{2} \bigr) \\ &\quad= E\bigl(Z^{2}\bigr)+E\bigl(Z_{1}^{2}\bigr)+E \bigl(Z_{2}^{2}\bigr)-2E\bigl(Z_{1}^{2} \bigr)-2E\bigl(Z_{2}^{2}\bigr)+ 2\bigl[E(Z) \bigr]^{2} \\ &\quad= E\bigl(Z^{2}\bigr)-E\bigl(Z_{1}^{2}\bigr)-E \bigl(Z_{2}^{2}\bigr)+ 2\bigl[E(Z)\bigr]^{2}. \end{aligned}$$
Combining with (7), we get
$$E\bigl(Z_{1}^{2}\bigr)-\bigl[E(Z)\bigr]^{2}\leq E \bigl(Z^{2}\bigr)- E\bigl(Z_{2}^{2}\bigr), $$
which is equivalent to the desired inequality. □
When X and Y are two finite discrete random variables, a discrete version of (6) is as follows:
$$ \sum_{i=1}^{m} \Biggl(\sum _{j=1}^{n}a_{ij}q_{j} \Biggr)^{2}p_{i}+\sum_{j=1}^{n} \Biggl(\sum_{i=1}^{m}a_{ij}p_{i} \Biggr)^{2}q_{j} \leq \Biggl(\sum _{i=1}^{m}\sum_{j=1}^{n}a_{ij}p_{i}q_{j} \Biggr)^{2}+ \sum_{i=1}^{m}\sum _{j=1}^{n}a_{ij}^{2}p_{i}q_{j}, $$
(8)
where the nonnegative real numbers \(p_{i}\) (\(1\leq i\leq m\)) and \(q_{j}\) (\(1\leq j\leq n\)) satisfy \(\sum_{i=1}^{m}p_{i}=1\) and \(\sum_{j=1}^{n}q_{i}=1\), \(A=(a_{ij})\) is a real \(m\times n\) matrix. This discrete version is given by Feng and Tonge [1].
If we put \(p_{i}=\frac{1}{m}\), \(i=1,2,\ldots,m\) and \(q_{j}=\frac {1}{n}\), \(j=1,2,\ldots,\frac{1}{n}\), then (8) becomes the following van Dam inequality:
$$ \sum_{i=1}^{m} \Biggl(\sum _{j=1}^{n}a_{ij} \Biggr)^{2} +\sum_{j=1}^{n} \Biggl( \sum_{i=1}^{m}a_{ij} \Biggr)^{2} \leq \Biggl(\sum_{i=1}^{m} \sum_{j=1}^{n}a_{ij} \Biggr)^{2}+ \sum_{i=1}^{m}\sum _{j=1}^{n}a_{ij}^{2}. $$
(9)
van Dam [2] applied his inequality to pose a related problem on the maximum irregularity of a directed graph with prescribed number of vertices and arcs.
In the special case of \((0,1)\)-matrices, the inequality (9) reduces to the following Khinchin-type inequality:
$$ m\sum_{i=1}^{m}r_{i}^{2} + n\sum_{j=1}^{n}c_{j}^{2} \leq\sigma^{2} +mn\sigma, $$
(10)
where \(r_{i}\), \(c_{i}\), and σ denote row sums, column sums, and entries summing of an \(m\times n\) \((0,1)\) matrix, respectively, as presented by Matúš and Tuzar [3]. This inequality is an improvement (in the nonsquare case) of a result by Khinchin [4], who proved that \(l \sum_{i=1}^{m} r_{i}^{2}+l \sum_{i=1}^{m} c_{i}^{2}\leq\sigma^{2}+l^{2}\sigma\), where \(l = \max\{m, n\}\). Khinchin [5] applied his inequality to prove a surprising number theoretic result.

Recently, Yan [6] presented another extension of (9). It is natural to ask whether the analog of (6) holds or not for three independent random variables. We have been unable to prove (or disprove) it.

Declarations

Acknowledgements

The many suggestions and detailed corrections of anonymous referees are gratefully acknowledged. Supported by the National Natural Science Foundation of China (11201039; 61273179).

Open Access This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors’ Affiliations

(1)
School of Information and Mathematics, Institute of Applied Mathematics, Yangtze University

References

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Copyright

© Yan and Zhang; licensee Springer. 2015