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Moment Inequalities and Jensen · Tutorial 385 of 994

Reversing Jensen With Concave Functions

Learn when concavity gives an upper bound on a transformed expectation and how to check the conditions before applying Jensen’s inequality.

Intermediate 9 min read

What You'll Learn

  • State the integrability and domain conditions for Jensen’s inequality with a concave function.
  • Determine why concavity reverses the direction of the Jensen bound.
  • Bound expectations of square roots and fractional powers using the mean of a nonnegative random variable.
  • Apply the logarithm bound when the random variable is positive and the logarithm is integrable.
  • Recognize common domain, integrability, and inequality-direction errors.

Why Concavity Gives an Upper Bound

A transformed expectation can be difficult to calculate even when the mean of the original random variable is known. Concavity provides a useful comparison: for a concave function, transforming the mean gives an upper bound on the mean of the transformed variable. This is the reverse direction from Jensen’s inequality for convex functions, and the direction depends on the shape of the function—not on a change in the definition of expectation.

For example, the square-root function is concave on the nonnegative real numbers. Thus, for a nonnegative random variable with finite mean, its expected square root cannot exceed the square root of its mean. This statement does not say the two quantities are usually equal; it gives a bound that can be computed using only the original mean.

Jensen’s Inequality for Concave Functions

Theorem: Let \(I\) be an interval, let \(X\) be a random variable such that \(P(X\in I)=1\), and suppose \(E[|X|]<\infty\). Let \(g:I\to\mathbb{R}\) be concave, with \(E[|g(X)|]<\infty\), and suppose \(E[X]\in I\). Then
$$ E[g(X)] \leq g(E[X]). $$

The integrability assumptions ensure that both sides are well-defined real numbers. The interval condition ensures that the function can be evaluated at both \(X\) and its mean. The essential feature is concavity: in the concave case, the transformed expectation lies below the transformation of the expectation.

One way to justify the direction is to apply the convex form of Jensen’s inequality to \(-g\). Since \(g\) is concave, \(-g\) is convex on the same interval. The assumptions above also ensure that \(E[|-g(X)|]=E[|g(X)|]<\infty\), so Jensen’s inequality applies:

$$ E[-g(X)] \geq -g(E[X]). $$

Multiplying both sides by \(-1\) reverses the inequality. Therefore \(E[g(X)]\leq g(E[X])\), as claimed. \(\square\)

For a twice-differentiable function on an interval, a common way to check concavity is to verify that \(g''(x)\leq 0\) throughout the interval. This test establishes the shape condition; it does not replace the need to check the random variable’s range and the relevant expectations.

Square Roots and Fractional Powers

For \(g(x)=\sqrt{x}\) on \([0,\infty)\), the function is continuous and concave. On \((0,\infty)\), its second derivative is

$$ g''(x)=-\frac{1}{4x^{3/2}}<0. $$

Concavity extends to the endpoint \(x=0\). Consequently, if \(X\geq 0\) almost surely and \(E[X]<\infty\), then \(E[\sqrt{X}]\) is finite: the pointwise inequality \(\sqrt{x}\leq 1+x\) for \(x\geq 0\) gives \(E[\sqrt{X}]\leq 1+E[X]\). Jensen’s inequality therefore yields the square-root bound.

Result: If \(X\geq 0\) almost surely and \(E[X]<\infty\), then
$$ E[\sqrt{X}] \leq \sqrt{E[X]}. $$

Worked Example: A Two-Point Square-Root Bound

Let \(X\) equal 1 with probability \(1/2\) and 9 with probability \(1/2\). Calculate the expected square root and compare it with the square root of the mean.

First compute the mean:

$$ E[X]=\frac{1}{2}(1)+\frac{1}{2}(9)=5. $$

The expected square root is

$$ E[\sqrt{X}] =\frac{1}{2}\sqrt{1}+\frac{1}{2}\sqrt{9} =\frac{1}{2}(1)+\frac{1}{2}(3) =2. $$

The transformed mean is \(\sqrt{E[X]}=\sqrt{5}\approx 2.236\). Thus \(2\leq \sqrt{5}\), in agreement with the concave Jensen bound. The two sides are not equal in this example.

More generally, for \(0<r<1\), the function \(g(x)=x^r\) is concave on \([0,\infty)\). For \(x>0\),

$$ g''(x)=r(r-1)x^{r-2}<0, $$

and the function is continuous at zero. If \(X\geq 0\) almost surely and \(E[X]<\infty\), then \(E[X^r]<\infty\): for \(x\geq 0\), \(x^r\leq 1+x\). Applying Jensen gives the following bound.

Result: If \(X\geq 0\) almost surely, \(E[X]<\infty\), and \(0<r<1\), then
$$ E[X^r]\leq (E[X])^r. $$

Worked Example: A Square-Root Bound for an Exponential Variable

Suppose \(X\) has an exponential distribution with rate \(2\), so its density is \(f(x)=2e^{-2x}\) for \(x>0\), and \(E[X]=1/2\). Find \(E[\sqrt{X}]\) and verify the Jensen bound.

Using the density,

$$ E[\sqrt{X}] =\int_0^\infty \sqrt{x}\,2e^{-2x}\,dx. $$

Substitute \(u=2x\), so \(x=u/2\) and \(dx=du/2\). Then

$$ E[\sqrt{X}] =\frac{1}{\sqrt{2}}\int_0^\infty u^{1/2}e^{-u}\,du =\frac{\Gamma(3/2)}{\sqrt{2}} =\frac{\sqrt{\pi}}{2\sqrt{2}} \approx 0.6267. $$

Meanwhile,

$$ \sqrt{E[X]}=\sqrt{\frac{1}{2}}\approx 0.7071. $$

The computed values satisfy \(0.6267\leq 0.7071\), as required. The integral is finite, so the expected square root exists and the bound applies.

Logarithms: Check Positivity and Integrability

The logarithm is concave on \((0,\infty)\), since \(g''(x)=-1/x^2<0\). Thus, if \(X>0\) almost surely, \(E[X]<\infty\), and \(E[|\log X|]<\infty\), Jensen’s inequality gives

$$ E[\log X]\leq \log(E[X]). $$

The positivity condition matters because \(\log X\) is not real-valued for nonpositive \(X\). The integrability condition matters because a finite mean for \(X\) alone does not guarantee that \(E[\log X]\) is a finite real number. Those are separate checks from concavity.

Worked Example: Comparing an Expected Logarithm With the Logarithm of the Mean

Let \(X=1\) with probability \(1/4\) and \(X=4\) with probability \(3/4\). Calculate the two sides of the logarithm bound.

The mean is

$$ E[X]=\frac{1}{4}(1)+\frac{3}{4}(4)=\frac{13}{4}=3.25. $$

Both possible values of \(X\) are positive, and \(X\) has finite support, so \(\log X\) is integrable. Its expectation is

$$ E[\log X] =\frac{1}{4}\log(1)+\frac{3}{4}\log(4) =\frac{3}{4}\log(4) \approx 1.0397. $$

The other side is

$$ \log(E[X])=\log(3.25)\approx 1.1787. $$

Therefore \(1.0397\leq 1.1787\), as Jensen’s inequality predicts. This calculation uses the logarithm only as a concave transformation to compare two expectations.

How to Apply the Bound Reliably

When using concavity to bound a transformed expectation, make the order of checks explicit. Doing so prevents an inequality with the right-looking expression from being applied on the wrong domain or in the wrong direction.

1
Identify the transformation.
Write the quantity of interest as \(E[g(X)]\), and identify the value \(g(E[X])\) that could provide a bound.
2
Check the domain.
Verify that \(X\) lies almost surely in an interval where \(g\) is defined and concave, and that \(E[X]\) belongs to that interval.
3
Check existence.
Ensure the expectations used in the inequality are well-defined; for the stated real-valued version, \(E[|X|]\) and \(E[|g(X)|]\) must be finite.
4
Apply the correct direction.
For concave \(g\), write \(E[g(X)]\leq g(E[X])\), then substitute the known mean and evaluate the right-hand side.

Common Mistakes and Why the Bound Matters

A frequent error is to remember the expression in Jensen’s inequality but forget which side is larger. Convexity gives \(g(E[X])\leq E[g(X)]\); concavity gives \(E[g(X)]\leq g(E[X])\). The function’s curvature determines the direction.

  • Using the convex direction for a concave function. For a square root, claiming \(\sqrt{E[X]}\leq E[\sqrt{X}]\) reverses the correct inequality. Check the sign of the second derivative on the relevant domain.
  • Applying the logarithm to a variable that might be nonpositive. Concavity cannot make an undefined real-valued expression valid. First establish \(X>0\) almost surely.
  • Assuming a finite mean is enough for every transformation. The function’s expectation must also exist under the version of Jensen being used. For the logarithm, verify integrability of \(\log X\), not just integrability of \(X\).
  • Confusing a bound with an equality. Jensen compares \(E[g(X)]\) and \(g(E[X])\); it does not assert they are interchangeable. Equality requires the relevant equality conditions, not merely that the function is concave.

The practical value of the result is that an upper bound for a transformed expectation can often be obtained without finding the full distribution of \(g(X)\). If the mean of \(X\) is known and the transformation is concave, the bound may require only one evaluation. It is a comparison, not generally a formula for the exact expectation.

Check Your Understanding

For each question, identify the domain and concavity considerations before writing the inequality.

  1. If \(X\geq 0\) almost surely and \(E[X]=16\), what upper bound does Jensen give for \(E[\sqrt{X}]\)?
  2. For \(0<r<1\), explain why \(E[X^r]\leq (E[X])^r\) applies when \(X\geq 0\) and \(E[X]<\infty\).
  3. A random variable takes values 2 and 8 with equal probabilities. Compute \(E[\log X]\) and \(\log(E[X])\), then check their order.
  4. Why does \(E[X]<\infty\) alone not establish that the real-valued logarithm version of Jensen’s inequality applies?
  5. For a concave function \(g\), which is the upper bound: \(E[g(X)]\) or \(g(E[X])\)?