Quick wins for a faster PC:
Clear out junk files and repair common Windows errorsFree Scan →Fix the driver behind crashes, sound loss and screen glitchesFind Drivers →Repair Windows errors before they cause bigger problemsFix Now →iTechGuides is reader-supported. When you buy through links on our site, we may earn an affiliate commission. As an Amazon Associate I earn from qualifying purchases. Learn more
For any two random variables with the needed moments, Var(XY) = E[X²Y²] − (E[XY])². This identity does not require independence. If X and Y are independent, it simplifies to Var(XY) = σX²σY² + σX²μY² + σY²μX², where μ denotes a mean and σ² a variance. The key is to use the simplified formula only when independence is justified.
The general formula works whether or not X and Y are independent
Set W = XY and apply the variance identity Var(W) = E[W²] − (E[W])². Since W² = X²Y², this gives:
Var(XY) = E[X²Y²] − (E[XY])².
This is the safest starting point. To evaluate it, you need the joint product moments E[XY] and E[X²Y²], and those moments must exist. Finite variances for X and Y individually do not, under dependence, automatically guarantee that E[X²Y²] is finite. See the Data 140 textbook’s variance and covariance properties.
The Tool Desk
Outbyte PC Repair FREEClear out junk files and repair common Windows errorsFree Scan →Outbyte Driver Updater FREEFix the driver behind crashes, sound loss and screen glitchesFind Drivers →For independent variables, marginal means and variances are enough
Let μX = E[X], μY = E[Y], σX² = Var(X), and σY² = Var(Y). If X and Y are independent, then E[XY] = μXμY and E[X²Y²] = E[X²]E[Y²]. Using E[X²] = σX² + μX² and E[Y²] = σY² + μY²:
#1 Best Overall
Var(XY) = (σX² + μX²)(σY² + μY²) − μX²μY²
Var(XY) = σX²σY² + σX²μY² + σY²μX².
Independence is what licenses both factorizations. The formula is not simply the product of the two variances: it also includes terms involving each variable’s mean. For a treatment of moments and independence, see the Georgia Tech-hosted probability textbook.
How the independent and dependent cases differ
| Case | What must be known | Can marginal means and variances alone suffice? |
|---|---|---|
| Independent X and Y | μX, μY, σX², and σY² | Yes. Independence lets the joint product moments factor. |
| Dependent X and Y | In general, E[XY] and E[X²Y²] from the joint distribution or another justified joint-moment model | No. Marginal means, variances, and covariance generally do not determine the product variance. |
Why covariance alone usually does not settle the dependent case
For dependent variables, covariance c = Cov(X,Y) describes their second-order association, but product variance can also depend on higher mixed moments. Write A = X − E[X] and B = Y − E[Y]. A centered expansion is:
Var(XY) = E[X]²Var(Y) + E[Y]²Var(X) + E[A²B²] + 2E[X]E[AB²] + 2E[Y]E[A²B] + 2E[X]E[Y]c − c².
Rank #3
- Brand New Textbook
- U.S Edition
- Fast shipping
The terms E[AB²], E[A²B], and E[A²B²] are third- and fourth-order mixed centered moments. Unless a model supplies these or directly supplies E[XY] and E[X²Y²], means, variances, and covariance are not generally enough. This issue is treated in Bohrnstedt and Goldberger’s paper, “On the Exact Covariance of Products of Random Variables”.
Check the assumptions before calculating
- Start with the general identity: use E[X²Y²] − (E[XY])².
- Use the independent shortcut only with independence: zero covariance alone is not the same condition as independence.
- Confirm the relevant moments exist: in particular, check that E[X²Y²] is finite if you need a finite variance.
- Do not confuse product variance with a product of variances: even under independence, mean-dependent terms remain.
When the two factors are the same variable
If Y = X, then XY = X², so Var(XY) = Var(X²) = E[X⁴] − (E[X²])². This case may require a fourth moment and is not covered by the independent-pair formula: a nonconstant variable is not independent of itself. The variance identity is also described in the Data 140 textbook.
Independent reader supportYour contribution helps us test, update, and keep practical guides available for everyone.For a product of several mutually independent variables
For mutually independent X₁,…,Xₙ, with μᵢ = E[Xᵢ] and σᵢ² = Var(Xᵢ), the corresponding formula is:
What’s actually slowing this PC down?
Pick the symptom - the matching free tool is one click away.
Var(∏ᵢ Xᵢ) = ∏ᵢ(σᵢ² + μᵢ²) − ∏ᵢ μᵢ².
Best Value
This follows by factoring the second moment of the product and its mean under mutual independence. It should not be carried over to dependent factors without accounting for their joint moments.
Quick Recap
Product prices and availability are accurate as of the date/time indicated and are subject to change. Any price and availability information displayed on Amazon at the time of purchase will apply.
