Jensen's inequality

In mathematics, Jensen's inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function. It was proved by Jensen in 1906, building on an earlier proof of the same inequality for doubly-differentiable functions by Otto Hölder in 1889. Given its generality, the inequality appears in many forms depending on the context, some of which are presented below. In its simplest form the inequality states that the convex transformation of a mean is less than or equal to the mean applied after convex transformation; it is a simple corollary that the opposite is true of concave transformations.

Jensen's inequality generalizes the statement that the secant line of a convex function lies ''above'' the graph of the function, which is Jensen's inequality for two points: the secant line consists of weighted means of the convex function (for ''t'' ∈  ,1,

:$t f(x_1) + (1-t) f(x_2),$

while the graph of the function is the convex function of the weighted means,

:$f(t x_1 + (1-t) x_2).$

Thus, Jensen's inequality is
:$f(t x_1 + (1-t) x_2) \leq t f(x_1) + (1-t) f(x_2).$

In the context of probability theory, it is generally stated in the following form: if ''X'' is a random variable and is a convex function, then

:\varphi(\operatorname \leq \operatorname \left varphi(X)\right

The difference between the two sides of the inequality, \operatorname \left varphi(X)\right- \varphi\left(\operatorname right), is called the Jensen gap.

Statements

The classical form of Jensen's inequality involves several numbers and weights. The inequality can be stated quite generally using either the language of measure theory or (equivalently) probability. In the probabilistic setting, the inequality can be further generalized to its ''full strength''.

Finite form

For a real convex function $\varphi$, numbers $x_1, x_2, \ldots, x_n$ in its domain, and positive weights $a_i$, Jensen's inequality can be stated as:

and the inequality is reversed if $\varphi$ is concave, which is

Equality holds if and only if $x_1=x_2=\cdots =x_n$ or $\varphi$ is linear on a domain containing $x_1,x_2,\cdots ,x_n$.

As a particular case, if the weights $a_i$ are all equal, then () and () become

For instance, the function is ''concave'', so substituting $\varphi(x) = \log(x)$ in the previous formula () establishes the (logarithm of the) familiar arithmetic-mean/geometric-mean inequality:

:\log\!\left( \frac\right) \geq \frac \quad \text \quad
\frac \geq \sqrt /math>

A common application has as a function of another variable (or set of variables) , that is, $x_i = g(t_i)$. All of this carries directly over to the general continuous case: the weights are replaced by a non-negative integrable function , such as a probability distribution, and the summations are replaced by integrals.

Measure-theoretic and probabilistic form

Let $(\Omega, A, \mu)$ be a probability space. Let $f : \Omega \to \mathbb$ be a $\mu$-measurable function and $\varphi : \mathbb \to \mathbb$ be convex. Then:
$\varphi\left(\int_\Omega f \,\mathrm\mu\right) \leq \int_\Omega \varphi \circ f \,\mathrm\mu$

In real analysis, we may require an estimate on

:$\varphi\left(\int_a^b f(x)\, dx\right)$

where $a, b \in \mathbb$, and f\colon , b\to \R is a non-negative Lebesgue- integrable function. In this case, the Lebesgue measure of , b/math> need not be unity. However, by integration by substitution, the interval can be rescaled so that it has measure unity. Then Jensen's inequality can be applied to get

:$\varphi\left(\frac\int_a^b f(x)\, dx\right) \le \frac \int_a^b \varphi(f(x)) \,dx.$

The same result can be equivalently stated in a probability theory setting, by a simple change of notation. Let $(\Omega, \mathfrak,\operatorname)$ be a probability space, ''X'' an integrable real-valued random variable and a convex function. Then:

:\varphi\left(\operatorname right) \leq \operatorname \left \varphi(X) \right

In this probability setting, the measure is intended as a probability $\operatorname$, the integral with respect to as an expected value $\operatorname$, and the function $f$ as a random variable ''X''.

Note that the equality holds if and only if is a linear function on some convex set $A$ such that $\mathrm(X \in A) = 1$ (which follows by inspecting the measure-theoretical proof below).

General inequality in a probabilistic setting

More generally, let ''T'' be a real topological vector space, and ''X'' a ''T''-valued integrable random variable. In this general setting, ''integrable'' means that there exists an element \operatorname /math> in ''T'', such that for any element ''z'' in the dual space of ''T'': $\operatorname, \langle z, X \rangle, <\infty$, and \langle z, \operatorname rangle = \operatorname langle z, X \rangle/math>. Then, for any measurable convex function and any sub- σ-algebra $\mathfrak$ of $\mathfrak$:

:\varphi\left(\operatorname\left \mid\mathfrak\rightright) \leq \operatorname\left varphi(X)\mid\mathfrak\right

Here \operatorname cdot\mid\mathfrak/math> stands for the expectation conditioned to the σ-algebra $\mathfrak$. This general statement reduces to the previous ones when the topological vector space is the real axis, and $\mathfrak$ is the trivial -algebra (where is the empty set, and is the sample space).

A sharpened and generalized form

Let ''X'' be a one-dimensional random variable with mean $\mu$ and variance $\sigma^2\ge 0$. Let $\varphi(x)$ be a twice differentiable function, and define the function

:$h(x)\triangleq\frac-\frac.$

Then

:
\sigma^2\inf \frac \le \sigma^2\inf h(x) \leq E\left varphi \left(X\right)\right\varphi\left(E right)\le \sigma^2\sup h(x) \le \sigma^2\sup \frac.

In particular, when $\varphi(x)$ is convex, then $\varphi''(x)\ge 0$, and the standard form of Jensen's inequality immediately follows for the case where $\varphi(x)$ is additionally assumed to be twice differentiable.

Proofs

Jensen's inequality can be proved in several ways, and three different proofs corresponding to the different statements above will be offered. Before embarking on these mathematical derivations, however, it is worth analyzing an intuitive graphical argument based on the probabilistic case where is a real number (see figure). Assuming a hypothetical distribution of values, one can immediately identify the position of \operatorname /math> and its image \varphi(\operatorname in the graph. Noticing that for convex mappings the corresponding distribution of values is increasingly "stretched out" for increasing values of , it is easy to see that the distribution of is broader in the interval corresponding to and narrower in for any ; in particular, this is also true for X_0 = \operatorname /math>. Consequently, in this picture the expectation of will always shift upwards with respect to the position of \varphi(\operatorname . A similar reasoning holds if the distribution of covers a decreasing portion of the convex function, or both a decreasing and an increasing portion of it. This "proves" the inequality, i.e.

:\varphi(\operatorname \leq \operatornamevarphi(X)= \operatorname

with equality when is not strictly convex, e.g. when it is a straight line, or when follows a degenerate distribution (i.e. is a constant).

The proofs below formalize this intuitive notion.

Proof 1 (finite form)

If and are two arbitrary nonnegative real numbers such that then convexity of implies

:$\forall x_1, x_2: \qquad \varphi \left (\lambda_1 x_1+\lambda_2 x_2 \right )\leq \lambda_1\,\varphi(x_1)+\lambda_2\,\varphi(x_2).$

This can be generalized: if are nonnegative real numbers such that , then

:$\varphi(\lambda_1 x_1+\lambda_2 x_2+\cdots+\lambda_n x_n)\leq \lambda_1\,\varphi(x_1)+\lambda_2\,\varphi(x_2)+\cdots+\lambda_n\,\varphi(x_n),$

for any .

The ''finite form'' of the Jensen's inequality can be proved by induction: by convexity hypotheses, the statement is true for ''n'' = 2. Suppose the statement is true for some ''n'', so

:$\varphi\left(\sum_^\lambda_i x_i\right) \leq \sum_^\lambda_i \varphi\left(x_i\right)$

for any such that .

One needs to prove it for . At least one of the is strictly smaller than $1$, say ; therefore by convexity inequality:

:$\begin \varphi\left(\sum_^\lambda_i x_i\right) &= \varphi\left((1-\lambda_)\sum_^ \frac x_i + \lambda_ x_ \right) \\ &\leq (1-\lambda_) \varphi\left(\sum_^ \frac x_i \right)+\lambda_\,\varphi(x_). \end$

Since ,

: $\sum_^ \frac = 1$,

applying the induction hypothesis gives

: $\varphi\left(\sum_^\frac x_i\right) \leq \sum_^\frac \varphi(x_i)$

therefore

: $\begin \varphi\left(\sum_^\lambda_i x_i\right) &\leq (1-\lambda_) \sum_^\frac \varphi(x_i)+\lambda_\,\varphi(x_) =\sum_^\lambda_i \varphi(x_i) \end$

We deduce the equality is true for , by the principle of mathematical induction it follows that the result is also true for all integer greater than 2.

In order to obtain the general inequality from this finite form, one needs to use a density argument. The finite form can be rewritten as:

:$\varphi\left(\int x\,d\mu_n(x) \right)\leq \int \varphi(x)\,d\mu_n(x),$

where ''μ''_''n'' is a measure given by an arbitrary convex combination of Dirac deltas:

:$\mu_n= \sum_^n \lambda_i \delta_.$

Since convex functions are continuous, and since convex combinations of Dirac deltas are weakly dense in the set of probability measures (as could be easily verified), the general statement is obtained simply by a limiting procedure.

Proof 2 (measure-theoretic form)

Let $g$ be a real-valued $\mu$-integrable function on a probability space $\Omega$, and let $\varphi$ be a convex function on the real numbers. Since $\varphi$ is convex, at each real number $x$ we have a nonempty set of subderivatives, which may be thought of as lines touching the graph of $\varphi$ at $x$, but which are at or below the graph of $\varphi$ at all points (support lines of the graph).

Now, if we define
:$x_0:=\int_\Omega g\, d\mu,$
because of the existence of subderivatives for convex functions, we may choose $a$ and $b$ such that
:$ax + b \leq \varphi(x),$
for all real $x$ and
:$ax_0+ b = \varphi(x_0).$
But then we have that
:$\varphi \circ g (\omega) \geq ag(\omega)+ b$
for almost all $\omega \in \Omega$. Since we have a probability measure, the integral is monotone with $\mu(\Omega) = 1$ so that

:$\int_\Omega \varphi \circ g\, d\mu \geq \int_\Omega (ag + b)\, d\mu = a\int_\Omega g\, d\mu + b\int_\Omega d\mu = ax_0 + b = \varphi (x_0) = \varphi \left (\int_\Omega g\, d\mu \right ),$

as desired.

Proof 3 (general inequality in a probabilistic setting)

Let ''X'' be an integrable random variable that takes values in a real topological vector space ''T''. Since $\varphi: T \to \R$ is convex, for any $x,y \in T$, the quantity

:$\frac,$

is decreasing as approaches 0⁺. In particular, the ''subdifferential'' of $\varphi$ evaluated at in the direction is well-defined by

:$(D\varphi)(x)\cdot y:=\lim_ \frac=\inf_ \frac.$

It is easily seen that the subdifferential is linear in (that is false and the assertion requires Hahn-Banach theorem to be proved) and, since the infimum taken in the right-hand side of the previous formula is smaller than the value of the same term for , one gets

:$\varphi(x)\leq \varphi(x+y)-(D\varphi)(x)\cdot y.$

In particular, for an arbitrary sub--algebra $\mathfrak$ we can evaluate the last inequality when x = \operatorname \mid\mathfrak\,y=X-\operatorname \mid\mathfrak/math> to obtain

:\varphi(\operatorname \mid\mathfrak \leq \varphi(X)-(D\varphi)(\operatorname \mid\mathfrak\cdot (X-\operatorname \mid\mathfrak.

Now, if we take the expectation conditioned to $\mathfrak$ on both sides of the previous expression, we get the result since:

:\operatorname \left [\left[(D\varphi)(\operatorname \mid\mathfrak\cdot (X-\operatorname \mid\mathfrak\right]\mid\mathfrak \right] = (D\varphi)(\operatorname \mid\mathfrak\cdot \operatorname[\left( X-\operatorname[X\mid\mathfrak] \right) \mid \mathfrak]=0,

by the linearity of the subdifferential in the ''y'' variable, and the following well-known property of the conditional expectation:

:\operatorname \left \left(\operatorname[X\mid\mathfrak\right)_\mid\mathfrak_\right_.html" ;"title="\mid\mathfrak.html" ;"title="\left(\operatorname[X\mid\mathfrak">\left(\operatorname[X\mid\mathfrak\right) \mid\mathfrak \right ">\mid\mathfrak.html" ;"title="\left(\operatorname[X\mid\mathfrak">\left(\operatorname[X\mid\mathfrak\right) \mid\mathfrak \right = \operatorname[ X \mid\mathfrak].

Applications and special cases

Form involving a probability density function

Suppose is a measurable subset of the real line and ''f''(''x'') is a non-negative function such that

:$\int_^\infty f(x)\,dx = 1.$

In probabilistic language, ''f'' is a probability density function.

Then Jensen's inequality becomes the following statement about convex integrals:

If ''g'' is any real-valued measurable function and $\varphi$ is convex over the range of ''g'', then

:$\varphi\left(\int_^\infty g(x)f(x)\, dx\right) \le \int_^\infty \varphi(g(x)) f(x)\, dx.$

If ''g''(''x'') = ''x'', then this form of the inequality reduces to a commonly used special case:

:$\varphi\left(\int_^\infty x\, f(x)\, dx\right) \le \int_^\infty \varphi(x)\,f(x)\, dx.$

This is applied in Variational Bayesian methods.

Example: even

moments of a random variable

If ''g''(''x'') = ''x²ⁿ'', and ''X'' is a random variable, then ''g'' is convex as
:$\frac(x) = 2n(2n - 1)x^ \geq 0\quad \forall\ x \in \R$
and so
:
g(\operatorname = (\operatorname ^ \leq\operatorname ^

In particular, if some even moment ''2n'' of ''X'' is finite, ''X'' has a finite mean. An extension of this argument shows ''X'' has finite moments of every order $l\in\N$ dividing ''n''.

Alternative finite form

Let and take to be the counting measure on , then the general form reduces to a statement about sums:

:$\varphi\left(\sum_^ g(x_i)\lambda_i \right) \le \sum_^ \varphi(g(x_i)) \lambda_i,$

provided that and

:$\lambda_1 + \cdots + \lambda_n = 1.$

There is also an infinite discrete form.

Statistical physics

Jensen's inequality is of particular importance in statistical physics when the convex function is an exponential, giving:

: e^ \leq \operatorname \left e^X \right

where the expected values are with respect to some probability distribution in the random variable .

Proof: Let $\varphi(x) = e^x$ in \varphi\left(\operatorname right) \leq \operatorname \left \varphi(X) \right

Information theory

If is the true probability density for , and is another density, then applying Jensen's inequality for the random variable and the convex function gives

:\operatorname varphi(Y)\ge \varphi(\operatorname

Therefore:

:$-D(p(x)\, q(x))=\int p(x) \log \left (\frac \right ) \, dx \le \log \left ( \int p(x) \frac\,dx \right ) = \log \left (\int q(x)\,dx \right ) =0$

a result called Gibbs' inequality.

It shows that the average message length is minimised when codes are assigned on the basis of the true probabilities ''p'' rather than any other distribution ''q''. The quantity that is non-negative is called the Kullback–Leibler divergence of ''q'' from ''p''.

Since is a strictly convex function for , it follows that equality holds when equals almost everywhere.

Rao–Blackwell theorem

If ''L'' is a convex function and $\mathfrak$ a sub-sigma-algebra, then, from the conditional version of Jensen's inequality, we get

:L(\operatorname delta(X) \mid \mathfrak \le \operatorname (\delta(X)) \mid \mathfrak\quad \Longrightarrow \quad \operatorname delta(X)_\mid_\mathfrak.html" ;"title="(\operatorname delta(X) \mid \mathfrak">(\operatorname delta(X) \mid \mathfrak\le \operatorname (\delta(X))

So if δ(''X'') is some estimator of an unobserved parameter θ given a vector of observables ''X''; and if ''T''(''X'') is a sufficient statistic for θ; then an improved estimator, in the sense of having a smaller expected loss ''L'', can be obtained by calculating

:\delta_1 (X) = \operatorname_ delta(X') \mid T(X')= T(X)

the expected value of δ with respect to θ, taken over all possible vectors of observations ''X'' compatible with the same value of ''T''(''X'') as that observed. Further, because T is a sufficient statistics, $\delta_1 (X)$ does not depend on θ, hence, becomes a statistics.

This result is known as the Rao–Blackwell theorem.

Financial Performance Simulation

A popular method of measuring the investment performance of an investment is the Internal Rate of Return (IRR) which is the rate by which a series of uncertain future cash flows are discounted using Present Value Theory to cause the sum of the future cash flows to equal the initial investment. While it is tempting to perform Monte Carlo simulation of the IRR, Jensen's Inequality introduces a bias due to fact that the IRR function is a curved function and the expectation operator is a linear function.

See also

* Karamata's inequality for a more general inequality
* Popoviciu's inequality
* Law of averages
* A proof without words of Jensen's inequality

Notes

References

*
* Tristan Needham (1993) "A Visual Explanation of Jensen's Inequality", American Mathematical Monthly 100(8):768–71.
*
*
*
*Sam Savage (2012
The Flaw of Averages: Why We Underestimate Risk in the Face of Uncertainty
(1st ed.) Wiley. ISBN 978-0471381976

External links

Jensen's Operator Inequality
of Hansen and Pedersen.
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{{Convex analysis and variational analysis

Convex analysis
Inequalities
Probabilistic inequalities
Statistical inequalities
Theorems in analysis
Theorems involving convexity
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