Common Errors in Applying Expectation Inequalities

Why Expectation Inequalities Need an Audit

Expectation inequalities are powerful because they turn information about moments into bounds on other quantities. But a familiar inequality is useful only when its assumptions match the random variables in the problem. Applying Markov's inequality to a signed variable, reversing Jensen's inequality for a convex function, or claiming equality without checking its conditions can produce a plausible-looking but false conclusion.

A reliable check has three parts: identify the inequality's hypotheses, verify its direction, and determine whether the claimed equality conditions actually hold. The examples below focus on Markov's inequality, Cauchy–Schwarz, and Jensen's inequality.

Markov's Inequality: Nonnegativity Is Essential

Markov's inequality: If $X$ is a nonnegative random variable with $E[X]<\infty$, then for every $t>0$, $$ P(X\ge t)\le \frac{E[X]}{t}. $$

The nonnegativity assumption is about the random variable being bounded, not merely about the threshold. For a signed variable $X$, it is generally not valid to replace $E[X]$ in this bound and claim $P(X\ge t)\le E[X]/t$. If $E[|X|]<\infty$, however, $|X|$ is nonnegative, and the event $\{X\ge t\}$ is contained in $\{|X|\ge t\}$. Markov's inequality therefore gives the valid bound $P(X\ge t)\le E[|X|]/t$.

The proof also shows precisely when Markov's inequality is an equality. Let $A=\{X\ge t\}$. Since $X\ge0$ almost surely and $X\ge t$ on $A$,

E[X]-tP(A) =E[(X-t)\mathbf{1}_A]+E[X\mathbf{1}_{A^c}] \ge 0.

Dividing by $t>0$ proves the bound. Equality holds exactly when both nonnegative terms on the right are zero: $X=t$ almost surely on $A$, and $X=0$ almost surely on $A^c$. Thus equality requires $X$ to take only the values $0$ and $t$, up to events of probability zero. A bound being valid does not mean it is attained.

Worked Example: Markov's Inequality Applied to a Signed Variable

Let $X=-1$ with probability $0.9$, and $X=9$ with probability $0.1$. A tempting but invalid calculation is to use $P(X\ge1)\le E[X]$, which is Markov's inequality with $t=1$.

The mean is $$ E[X]=(-1)(0.9)+(9)(0.1)=-0.9+0.9=0. $$ But $P(X\ge1)=0.1$, so the proposed bound would say $0.1\le0$, which is false. The failure is that $X$ is not nonnegative.

A valid repair uses $|X|$. Here $$ E[|X|]=(1)(0.9)+(9)(0.1)=0.9+0.9=1.8, $$ and $\{X\ge1\}\subseteq\{|X|\ge1\}$. Markov's inequality gives $P(X\ge1)\le1.8$, which is true but not sharp. The direct probability, $0.1$, is much smaller; a valid bound need not be informative in every case.

Cauchy–Schwarz: Use Second Moments, Not Products of Means

For square-integrable real random variables, Cauchy–Schwarz controls the expectation of a product using second moments. It does not, without further assumptions, identify that expectation with the product of the means.

Cauchy–Schwarz inequality: If $E[X^2]<\infty$ and $E[Y^2]<\infty$, then $$ |E[XY]|\le \sqrt{E[X^2]E[Y^2]}. $$ If both second moments are positive, equality holds exactly when $X=cY$ almost surely for some constant $c$.

For completeness, if $E[Y^2]>0$, then for every real $s$, $$ 0\le E[(X-sY)^2] =E[X^2]-2sE[XY]+s^2E[Y^2]. $$ This quadratic in $s$ cannot be negative at any real value. Its discriminant must therefore be nonpositive: $$ 4(E[XY])^2-4E[X^2]E[Y^2]\le0. $$ Taking square roots gives the inequality. If $E[Y^2]=0$, then $Y=0$ almost surely, so $E[XY]=0$ and the result also holds. When both second moments are positive, equality requires the quadratic to be zero at its minimum, which is equivalent to $X=sY$ almost surely for $s=E[XY]/E[Y^2]$. $\square$

A frequent error is to replace $E[X^2]$ by $(E[X])^2$, or to replace $E[XY]$ by $E[X]E[Y]$, without a reason that justifies the replacement. Cauchy–Schwarz concerns second moments and does not require independence. The identity $E[XY]=E[X]E[Y]$, by contrast, is guaranteed by independence when the relevant expectations exist; it does not follow from Cauchy–Schwarz.

Worked Example: A Product Expectation Is Not a Product of Means

Let $X=Y$, where $X$ is Bernoulli with $P(X=1)=0.2$ and $P(X=0)=0.8$. Then $X$ and $Y$ are dependent because they are the same random variable.

Their means and product expectation are $$ E[X]=E[Y]=0.2,\qquad E[XY]=E[X^2]=0.2. $$ Thus $E[XY]=0.2$, whereas $E[X]E[Y]=(0.2)(0.2)=0.04$. Treating these as equal would be incorrect.

Cauchy–Schwarz gives the valid calculation $$ |E[XY]|=0.2 \le \sqrt{E[X^2]E[Y^2]} =\sqrt{(0.2)(0.2)}=0.2. $$ Equality holds because $X=Y$ almost surely, so the variables are proportional. This example also shows why the second moments in Cauchy–Schwarz cannot be casually replaced by squared means.

Jensen's Inequality: Direction and Equality Depend on the Function

For Jensen's inequality, first identify whether the function is convex or concave on the range of the random variable. That classification determines the direction; the fact that an expectation is being bounded does not.

Jensen's inequality: Let $I$ be an interval, let $\phi:I\to\mathbb{R}$ be convex, and suppose $X\in I$ almost surely, $E[|X|]<\infty$, $E[X]\in I$, and $E[|\phi(X)|]<\infty$. Then $$ \phi(E[X])\le E[\phi(X)]. $$ For a concave $\phi$, the inequality is reversed, under the same integrability and domain conditions.

The condition $E[|X|]<\infty$ ensures that $E[X]$ exists as a finite mean. The condition on $\phi(X)$ ensures that the expectation on the other side is also finite. In some settings Jensen's inequality has an extended-expectation version, but one must then check its assumptions rather than perform ordinary finite arithmetic with a nonexistent or infinite expectation.

Another common mistake is to infer equality merely because $X$ is bounded or because the inequality appears numerically close. For a strictly convex function, equality under the stated conditions requires $X$ to be constant almost surely. For a convex function that is not strictly convex, a nonconstant $X$ can give equality when the function is affine over the values taken by $X$.

Worked Example: Reversing Jensen's Inequality for an Exponential

Let $X$ take the values $0$ and $2$, each with probability $1/2$, and set $\phi(x)=e^x$. Since $\phi''(x)=e^x>0$, the function is strictly convex. A mistaken reversal would claim $E[e^X]\le e^{E[X]}$.

The mean is $$ E[X]=\frac{0+2}{2}=1, $$ and the two sides are $$ E[e^X]=\frac{e^0+e^2}{2}=\frac{1+e^2}{2}\approx 4.195, \qquad e^{E[X]}=e\approx2.718. $$ The claimed reversed inequality would say $4.195\le2.718$, which is false. Convexity gives the correct direction: $$ e^{E[X]}\le E[e^X]. $$ The inequality is strict because $X$ is nonconstant and the exponential function is strictly convex.

Worked Example: Equality for a Nonconstant Variable

Let $X$ take the values $1$ and $3$, each with probability $1/2$, and define $\phi(x)=\max(x,0)$. This function is convex but not strictly convex: on the positive half-line it equals the affine function $x$.

Here $E[X]=(1+3)/2=2$, and both possible values of $X$ are positive. Consequently, $$ E[\phi(X)]=\frac{\phi(1)+\phi(3)}{2} =\frac{1+3}{2}=2 =\phi(2)=\phi(E[X]). $$ The variable is not constant, yet equality holds. Therefore the claim “equality in Jensen implies $X$ is constant” is too broad: that conclusion requires strict convexity (with the relevant integrability conditions), not merely convexity.

A Practical Audit Before Using a Bound

Before relying on an expectation inequality, write down the object to which it is being applied and check the assumptions against that object. A compact audit helps separate a valid bound from a superficially similar invalid one.

Check existence.
Verify that the expectations used in the calculation exist in the required sense. Do not manipulate an undefined mean as if it were a finite number.

Check the variable's properties.
Markov requires a nonnegative integrable variable and a positive threshold. Cauchy–Schwarz requires finite second moments. Jensen requires a finite mean, a suitable domain, and the needed integrability of the function of the variable.

Check the direction.
Convexity gives $ \phi(E[X])\le E[\phi(X)] $; concavity reverses it. Do not reverse a bound without a condition that actually licenses the reversal.

Check equality separately.
State the equality conditions for the particular inequality. A correct inequality can still have a false equality claim.

Common Errors

Applying Markov's inequality to a signed variable. The required nonnegative quantity may be $|X|$, not $X$. Verify that the event of interest is contained in the event to which Markov is applied.
Confusing a bound with an identity. Cauchy–Schwarz bounds $|E[XY]|$; it does not say $E[XY]=E[X]E[Y]$.
Reversing Jensen's inequality by intuition. Determine convexity or concavity on the relevant interval before writing the inequality.
Claiming equality from convexity alone. Strict convexity is important for the usual constant-variable equality conclusion; an affine portion can permit equality for a nonconstant variable.
Ignoring integrability. A formula involving $E[X]$, $E[X^2]$, or $E[\phi(X)]$ must be supported by the existence conditions required for that formula.

These checks matter in applications because a moment bound is only as reliable as the reasoning that produced it. In particular, a weak but valid inequality is preferable to a sharper-looking statement that silently assumes nonnegativity, independence, strict convexity, or finite expectations.

Check Your Understanding

For each question, identify the relevant assumption or equality condition before calculating.

Why can Markov's inequality be applied to $|X|$ when $E[|X|]<\infty$, even if $X$ takes negative values?
Suppose $E[X^2]$ and $E[Y^2]$ are finite. State the Cauchy–Schwarz bound and the condition for equality when both second moments are positive.
If $\phi$ is concave and the expectations exist, which direction does Jensen's inequality take?
Can equality in Jensen's inequality for a convex function hold when $X$ is nonconstant? Give the condition illustrated by the worked example.
What goes wrong if a proof applies Markov's inequality directly to a signed $X$ and substitutes $E[X]$ for the nonnegative expectation?