Open Access

On a probability inequality of van Dam

Journal of Inequalities and Applications20152015:128

Received: 3 July 2014

Accepted: 31 March 2015

Published: 11 April 2015


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).


probability inequality random variable van Dam’s inequality


Let T be a continuous random variable having the probability density function ϕ defined on R. By definition
$$ E(T)=\int_{-\infty}^{\infty} t\phi(t)\,dt $$
is the expectation of T, and
$$ \sigma^{2}(T)=\int_{-\infty}^{\infty} \bigl(t-E(T)\bigr)^{2}\phi(t)\,dt $$
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
$$\begin{aligned}& \begin{aligned}[b] E\bigl(E(Z|X)\bigr) &= \int_{-\infty}^{\infty} \biggl(\int _{-\infty}^{\infty} f(x,y) \phi_{2}(y)\,dy \biggr) \phi_{1}(x)\,dx\\ & = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) \phi_{2}(y)\phi_{1}(x)\,dx \,dy =E(Z), \end{aligned}\\& \begin{aligned}[b] E\bigl(Z \cdot E(Z|X)\bigr) &= \int_{-\infty}^{\infty}\int _{-\infty}^{\infty} f(x,y) \biggl(\int_{-\infty}^{\infty} f(x,t) \phi _{2}(t)\,dt \biggr) \phi_{1}(x) \phi_{2}(y) \,dx\,dy\\ & = \int_{-\infty}^{\infty} \biggl(\int _{-\infty}^{\infty} f(x,y) \phi_{2}(y)\,dy \biggr)^{2}\phi_{1}(x)\,dx =E\bigl(\bigl[E(Z|X) \bigr]^{2}\bigr), \end{aligned} \end{aligned}$$
$$\begin{aligned} &E\bigl(E(Z|X)\cdot E(Z|Y)\bigr) \\ &\quad= \int_{-\infty}^{\infty }\int_{-\infty}^{\infty} \biggl(\int_{-\infty}^{\infty} f(s,y) \phi_{1}(s)\,ds \biggr) \biggl(\int_{-\infty}^{\infty} f(x,t) \phi_{2}(t)\,dt \biggr) \phi_{1}(x)\phi _{2}(y) \,dx\,dy \\ &\quad = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(s,y) \phi_{1}(s) \phi_{2}(y)\,ds \,dy \int_{-\infty}^{\infty} \int_{-\infty }^{\infty} f(x,t) \phi_{1}(x) \phi_{2}(t)\,dx \,dt =\bigl[E(Z)\bigr]^{2}. \end{aligned}$$
$$\begin{aligned}& E\bigl(E(Z|X)\bigr)= E(Z), \end{aligned}$$
$$\begin{aligned}& E\bigl(Z \cdot E(Z|X)\bigr) = E\bigl(\bigl[E(Z|X ) \bigr]^{2}\bigr), \end{aligned}$$
$$\begin{aligned}& E\bigl(E(Z|X)\cdot E(Z|Y)\bigr) = \bigl[E(Z) \bigr]^{2}. \end{aligned}$$

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
$$ E\bigl(\sigma^{2}(Z|X)\bigr)\leq\sigma^{2} \bigl(E(Z|Y)\bigr). $$


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 \(\sigma^{2}(Z-Z_{1}-Z_{2})=E([Z-Z_{1}-Z_{2}]^{2})-[E(Z-Z_{1}-Z_{2})]^{2}\geq0\),
$$ \bigl[E(Z)\bigr]^{2}\leq E\bigl([Z-Z_{1}-Z_{2}]^{2} \bigr). $$
From (4) and (5), we have
$$E(ZZ_{1}) =E\bigl(Z_{1}^{2}\bigr),\qquad E(ZZ_{2})=E\bigl(Z_{2}^{2}\bigr) $$
$$E(Z_{1}Z_{2}) = \bigl[E(Z)\bigr]^{2}. $$
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:
$$ \sum_{i=1}^{m} \Biggl(\sum _{j=1}^{n}a_{ij}q_{j} \Biggr)^{2}p_{i}+\sum_{j=1}^{n} \Biggl(\sum_{i=1}^{m}a_{ij}p_{i} \Biggr)^{2}q_{j} \leq \Biggl(\sum _{i=1}^{m}\sum_{j=1}^{n}a_{ij}p_{i}q_{j} \Biggr)^{2}+ \sum_{i=1}^{m}\sum _{j=1}^{n}a_{ij}^{2}p_{i}q_{j}, $$
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:
$$ \sum_{i=1}^{m} \Biggl(\sum _{j=1}^{n}a_{ij} \Biggr)^{2} +\sum_{j=1}^{n} \Biggl( \sum_{i=1}^{m}a_{ij} \Biggr)^{2} \leq \Biggl(\sum_{i=1}^{m} \sum_{j=1}^{n}a_{ij} \Biggr)^{2}+ \sum_{i=1}^{m}\sum _{j=1}^{n}a_{ij}^{2}. $$
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:
$$ m\sum_{i=1}^{m}r_{i}^{2} + n\sum_{j=1}^{n}c_{j}^{2} \leq\sigma^{2} +mn\sigma, $$
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.



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

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Authors’ Affiliations

School of Information and Mathematics, Institute of Applied Mathematics, Yangtze University


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© Yan and Zhang; licensee Springer. 2015