Why Independence Implies Zero Covariance

Why Independence Gives Zero Covariance

Covariance measures whether two random variables tend to vary together. Independence is a stronger condition: knowing one variable does not change the distribution of the other. That stronger condition has a direct consequence for covariance. When the required expectations exist, independence lets us factor the expectation of a product into the product of expectations. Subtracting those equal quantities gives zero covariance.

The result is useful in actuarial calculations because it can simplify expressions involving independent counts, indicators, or losses. The key is to check the conditions before using it: the variables must actually be independent, and their expectations must exist.

Definition: For random variables \(X\) and \(Y\) with finite means and a finite product expectation, their covariance is

\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y].

If \(X\) and \(Y\) are independent and have finite first absolute moments, then \(E[XY]=E[X]E[Y]\). Consequently, \(\operatorname{Cov}(X,Y)=0\).

The finite-moment condition matters. For example, the formula for covariance requires that the means and product expectation be defined and finite; it is not legitimate to subtract undefined or infinite quantities. A sufficient condition for the result is \(E[|X|]<\infty\) and \(E[|Y|]<\infty\). Independence then also gives \(E[|XY|]=E[|X|]E[|Y|]<\infty\).

Why the Product Expectation Factors

Independence means that the joint distribution is the product of the marginal distributions. In the discrete case, if \(X\) and \(Y\) have probability mass functions \(p_X\) and \(p_Y\), independence gives \(P(X=x,Y=y)=p_X(x)p_Y(y)\). Summing over all possible pairs separates the two sums:

\begin{aligned} E[XY] &=\sum_x\sum_y xy\,P(X=x,Y=y)\\ &=\sum_x\sum_y xy\,p_X(x)p_Y(y)\\ &=\left(\sum_x x\,p_X(x)\right) \left(\sum_y y\,p_Y(y)\right)\\ &=E[X]E[Y]. \end{aligned}

The same idea applies when the variables have a joint density. If \(f_{X,Y}(x,y)=f_X(x)f_Y(y)\), then, whenever the expectations are integrable,

\begin{aligned} E[XY] &=\int\!\!\int xy f_{X,Y}(x,y)\,dx\,dy\\ &=\int\!\!\int xy f_X(x)f_Y(y)\,dx\,dy\\ &=\left(\int x f_X(x)\,dx\right) \left(\int y f_Y(y)\,dy\right)\\ &=E[X]E[Y]. \end{aligned}

In either case, substituting the factorization into the definition of covariance completes the argument:

\operatorname{Cov}(X,Y) =E[XY]-E[X]E[Y] =E[X]E[Y]-E[X]E[Y] =0.

Theorem: If \(X\) and \(Y\) are independent and \(E[|X|]\) and \(E[|Y|]\) are finite, then their covariance exists and equals zero. Independence is the condition that permits the product-expectation factorization.

Worked Examples

Worked Example: Independent Indicators with Affine Transformations

In a hypothetical portfolio, let \(X\) indicate whether a randomly selected policy has a specified feature, with \(P(X=1)=0.3\). Let \(Y\) indicate whether an independent review event occurs, with \(P(Y=1)=0.4\). Define \(A=2+3X\) and \(B=5-2Y\). Find \(\operatorname{Cov}(A,B)\).

Method and assumptions. The indicators \(X\) and \(Y\) are independent Bernoulli variables with parameters \(0.3\) and \(0.4\). The variables \(A\) and \(B\) are functions of \(X\) and \(Y\), respectively, so they are independent as well. All variables have finite support, so the required expectations exist.

First calculate the means of the indicators:

E[X]=0(0.7)+1(0.3)=0.3,\qquad E[Y]=0(0.6)+1(0.4)=0.4.

Then calculate the means of the transformed variables:

E[A]=2+3E[X]=2+3(0.3)=2.9, \qquad E[B]=5-2E[Y]=5-2(0.4)=4.2.

Because \(A\) and \(B\) are independent, factor the product expectation:

$$ E[AB]=E[A]E[B]=(2.9)(4.2)=12.18. $$

Therefore,

\operatorname{Cov}(A,B)=12.18-(2.9)(4.2)=12.18-12.18=0.

Check. The product expectation equals the product of the means, as required. This method avoids calculating all four possible pairs of indicator outcomes separately.

Worked Example: A Continuous Variable and a Squared Variable

Let \(X\sim\operatorname{Uniform}(0,4)\) and let \(Y\) have an exponential distribution with rate \(0.5\), so its mean is \(1/0.5=2\). Assume \(X\) and \(Y\) are independent. Find \(\operatorname{Cov}(X^2,Y)\), including \(E[X^2Y]\).

Method and assumptions. Both variables have finite moments needed here. Independence of \(X\) and \(Y\) implies independence of the functions \(X^2\) and \(Y\), so \(E[X^2Y]=E[X^2]E[Y]\).

The density of \(X\) is \(1/4\) on \(0<x<4\). Thus,

\begin{aligned} E[X^2] &=\int_0^4 x^2\frac14\,dx\\ &=\frac14\left[\frac{x^3}{3}\right]_0^4\\ &=\frac14\left(\frac{64}{3}\right) =\frac{16}{3}. \end{aligned}

For an exponential variable with rate \(0.5\), \(E[Y]=1/0.5=2\). Factor the product expectation:

E[X^2Y]=E[X^2]E[Y] =\frac{16}{3}(2)=\frac{32}{3}.

Also, \(E[X^2]E[Y]=(16/3)(2)=32/3\), so

\operatorname{Cov}(X^2,Y) =E[X^2Y]-E[X^2]E[Y] =\frac{32}{3}-\frac{32}{3}=0.

Check. The calculation uses the second moment of \(X\), not \((E[X])^2\). The covariance is zero because \(X^2\) and \(Y\) remain independent.

Worked Example: Verifying the Result from a Joint Probability Table

Let \(X\) take values \(0\) and \(2\), with probabilities \(0.25\) and \(0.75\). Independently, let \(Y\) take values \(-1\) and \(3\), each with probability \(0.5\). Find \(E[XY]\) and \(\operatorname{Cov}(X,Y)\).

Method and assumptions. The variables are independent and have finite support. Their joint probabilities are products of the marginal probabilities. In particular, each joint probability in the table below is the probability of the corresponding pair.

	\(Y=-1\)	\(Y=3\)
\(X=0\)	\((0.25)(0.5)=0.125\)	\((0.25)(0.5)=0.125\)
\(X=2\)	\((0.75)(0.5)=0.375\)	\((0.75)(0.5)=0.375\)

The marginal means are

E[X]=0(0.25)+2(0.75)=1.5, \qquad E[Y]=(-1)(0.5)+3(0.5)=1.

Directly summing \(xy\) times each joint probability gives

\begin{aligned} E[XY] &=(0)(-1)(0.125)+(0)(3)(0.125)\\ &\quad +(2)(-1)(0.375)+(2)(3)(0.375)\\ &=0+0-0.75+2.25\\ &=1.5. \end{aligned}

Therefore,

\operatorname{Cov}(X,Y)=1.5-(1.5)(1)=0.

Check. The direct joint sum gives \(E[XY]=1.5\), which agrees with \(E[X]E[Y]=(1.5)(1)=1.5\). This verifies the factorization numerically.

Common Mistake: Factoring Without Independence

The identity \(E[XY]=E[X]E[Y]\) is not a general rule for every pair of random variables. It is justified here by independence. A tempting but invalid calculation is to replace a joint expectation with a product of marginal expectations merely because both marginal distributions are known.

For a concrete check, let \(X=Y\), where \(X\) is Bernoulli with parameter \(0.5\). Then \(E[X]=E[Y]=0.5\). The incorrect factorization would give \(E[XY]=(0.5)(0.5)=0.25\). But \(XY=X^2=X\), since \(X\) is either 0 or 1, so the actual product expectation is \(E[XY]=0.5\). The covariance is therefore

\operatorname{Cov}(X,Y)=0.5-(0.5)(0.5)=0.25,

not zero. The correction is to establish independence before factoring. In this example \(X\) and \(Y\) are the same variable, so they are not independent.

Check the model.
Identify whether the variables are independent; do not infer independence just from knowing both marginal distributions.

Check existence.
Confirm the means and product expectation are finite, or use finite first absolute moments as a sufficient condition.

Factor and subtract.
For independent variables, use \(E[XY]=E[X]E[Y]\), then apply the covariance definition to obtain zero.

Using the Result Correctly

The theorem is especially efficient when the joint distribution would otherwise require a long sum or double integral. Once independence and existence are established, there is no need to calculate the full joint distribution to find covariance. Conversely, if independence is not given or cannot be established, the product expectation must be handled using the available joint information.

A zero covariance is a statement about the expected centered product, not a proof that the variables are independent. The result established here is the forward implication: under the stated moment conditions, independence implies zero covariance. It should not be reversed without additional assumptions.

Key takeaway: For independent \(X\) and \(Y\) with finite required expectations, independence factors the joint product expectation:

E[XY]=E[X]E[Y] \quad\Longrightarrow\quad \operatorname{Cov}(X,Y)=0.

The most important exam check is whether independence and the moment conditions are actually satisfied.

Check Your Understanding

For each problem, state the relevant independence or dependence condition and compute the requested quantity.

Independent Bernoulli variables \(X\) and \(Y\) have parameters \(0.2\) and \(0.6\). Let \(A=1+10X\) and \(B=-2+4Y\). Find \(E[A]\), \(E[B]\), \(E[AB]\), and \(\operatorname{Cov}(A,B)\). (Answer: \(E[A]=3\), \(E[B]=0.4\), \(E[AB]=1.2\), covariance \(=0\).)
Independent variables satisfy \(X\sim\operatorname{Uniform}(0,3)\) and \(Y\sim\operatorname{Exponential}(1)\), where the exponential parameter is the rate. Find \(E[X^2]\), \(E[5Y+1]\), \(E[X^2(5Y+1)]\), and \(\operatorname{Cov}(X^2,5Y+1)\). (Answer: \(3\), \(6\), \(18\), and \(0\), respectively.)
Independent \(X\) and \(Y\) have \(P(X=-2)=0.25\), \(P(X=1)=0.75\), \(P(Y=0)=0.5\), and \(P(Y=4)=0.5\). Find \(E[X]\), \(E[Y]\), \(E[XY]\), and the covariance. (Answer: \(E[X]=0.25\), \(E[Y]=2\), \(E[XY]=0.5\), covariance \(=0\).)
Suppose \(X=Y\) and \(X\) is Bernoulli with parameter \(0.3\). Calculate \(E[X]E[Y]\), \(E[XY]\), and \(\operatorname{Cov}(X,Y)\). Explain why independence-based factorization is unavailable. (Answer: \(0.09\), \(0.3\), and \(0.21\); the variables are identical and are not independent.)
Independent variables \(X\) and \(Y\) have finite first absolute moments \(E[|X|]=2\) and \(E[|Y|]=3\). What is \(E[|XY|]\), and what does this imply about the integrability needed for the zero-covariance result? (Answer: \(E[|XY|]=6\), so the product is absolutely integrable; independence gives \(E[XY]=E[X]E[Y]\) and covariance \(0\).)

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