Measure dispersion and recover moments from tails

Try

Two variables can share a mean while differing sharply in spread. Suppose a model has E(X)=1 and E(X²)=5. The mean squared distance from one is 5−1=4. Explain why the second moment alone is not variance unless the mean is zero.

Predict

Predict how variance changes when Y=7−3X. Adding seven shifts the center, while multiplying by negative three reverses direction and triples distances. Squared distances therefore scale by nine. A variance cannot become negative simply because a scale multiplier is negative.

Experiment

For the model above, Var(Y)=9·4=36. Verify the same principle with a two-point law: X is 0 or 4 equally likely, with mean 2 and variance 4. Y then takes 7 or −5, with mean 1 and squared deviations 36 in either case.

Draw a rectangle representation of a nonnegative observed value x: x is the area under the indicator 1(t<x) for t≥0. For x², replace the rectangle height by 2t and integrate from zero to x. These pointwise areas lead to tail formulas after averaging over X.

Understand

With finite second moment, variance is Var(X)=E[(X−μ)²]=E(X²)−μ², where μ=E(X). Finite E(X²) also ensures a finite first absolute moment by Cauchy–Schwarz. The expanded formula is safe under those conditions; subtracting infinite moments is not a valid way to assign a finite variance.

Variance is nonnegative. If it is zero, the nonnegative variable (X−μ)² has expectation zero and therefore vanishes almost surely. This means X=μ outside a probability-zero set, not necessarily at every abstract sample-space point. Standard deviation is the square root and has the original measurement units; variance has squared units.

For an affine transformation a+bX, variance is b²Var(X). The shift disappears after centering. This formula does not require independence because only one variable is involved. Adding two distinct variables requires covariance information, developed later.

For nonnegative X, the tail integral E(X)=∫₀∞P(X>t)dt follows from X=∫₀∞1(t<X)dt. Tonelli’s theorem permits interchanging expectation and integration because the integrand is nonnegative, even when the answer is infinite. Similarly E(X²)=∫₀∞2tP(X>t)dt. Higher positive powers have analogous formulas under appropriate exponents.

For a positive integer-valued variable, the corresponding area becomes a sum of survival probabilities. Keeping the index convention matters: E(G)=Σ from k=0 to infinity P(G>k) for support starting at one. A shifted support changes how the formula is written, not the underlying counting of layers.

These identities connect tail decay to moment existence. A distribution can have a finite mean but infinite variance; faster decay may be needed for higher moments. Tail-based arguments can reveal this without finding a complicated density antiderivative. A finite variance is an assumption to check in limit theorems, not an automatic property of every random variable.

Apply

Derive Uniform(0,a) moments directly and confirm the variance has squared-length units if a is a length. Then give the tail-integral proof with its nonnegativity justification. Do not cite an interchange theorem without explaining which hypothesis makes the interchange valid.

Continue

Next derive transformed distributions themselves. Unlike LOTUS, a density transformation must account for every inverse branch and for any mass collapsed to a point.