Distribution tails: the same moments, different worst days

Two distributions can share the same mean and the same variance and still carry very different worst days. The moments you quote do not pin the tail.

Live result

A places 0% outside the threshold and B places 20%, on the same mean and the same variance.

threshold1.5
mean of both distributions0
variance of both distributions1
tail difference in points20

Equal moments do not settle what sits out in the tail.

Shading marks values with magnitude greater than 1.5; equality is excluded.

The moments match

Distribution A

Mean: 0
: 1
Outside threshold: 0%

Distribution B

Mean: 0
: 1
Outside threshold: 20%

Support and tail ledger
Distribution	Value	Exact probability	Outside threshold
A	-1	$\frac{1}{2}$	No
A	1	$\frac{1}{2}$	No
B	-0.5	$\frac{4}{5}$	No
B	2	$\frac{1}{5}$	Yes

The deciding record

Tail probability gap: 20 percentage points
Mean 0 and variance 1 stayed fixed: Mean 0 and variance 1 stayed fixed
At $k = 1.5$ , neither ±1 lies outside. The value +2 does, carrying probability $\frac15 = 20\%$ .: $P(\lvert A\rvert > k)=0,\qquad P(\lvert B\rvert > k)=\frac15,\qquad 1<k<2$ $\begin{gathered}P(\lvert A\rvert > k)=0, \\ P(\lvert B\rvert > k)=\frac15, \\ 1<k<2\end{gathered}$

Boundary: Matching two moments alone does not identify a distribution or estimate a real tail.

Boundary and source

Source: MIT Probability and Random Variables grounds the discrete probability calculation. NIST's normal distribution reference supports the repaired statement that mean and variance identify the member only after a suitable kind of distribution is fixed. No source wording is reproduced.

Open the moment check

The exact two-distribution witness

Scale: 1
Distribution A: −1 and +1, each with probability $\frac12$
Distribution B: $-\frac12$ with probability $\frac45$ , and +2 with probability $\frac15$
Tail rule: Count support values whose absolute value is strictly greater than the threshold

At $k = 1.5$ , neither ±1 lies outside. The value +2 does, carrying probability $\frac15 = 20\%$ .

The equal moments

For A, the equal weights at −1 and +1 cancel to mean zero and give variance one. For B,

(-\frac12\times\frac45)+(2\times\frac15)=0

, while

((\frac12)^2\times\frac45)+(2^2\times\frac15)=1

P(\lvert A\rvert > k)=0,\qquad P(\lvert B\rvert > k)=\frac15,\qquad 1<k<2

\begin{gathered}P(\lvert A\rvert > k)=0, \\ P(\lvert B\rvert > k)=\frac15, \\ 1<k<2\end{gathered}

where $P(\lvert A\rvert>k)$ is A's tail probability and $P(\lvert B\rvert>k)$ B's tail probability, for thresholds between $1$ and $2$ .

Worked example

At $k = 1.5$ , the values −1 and +1 of A are both inside the threshold, so its tail probability is 0. In B, −0.5 is inside and +2 is outside. The outside point carries probability $\frac15$ , so B's tail probability is 0.2. Both distributions still have mean 0 and variance 1.

Between one and two there is no support point in either distribution except B's value at plus two, so every admitted threshold returns the same twenty point gap.

This construction does not estimate real tail risk, identify a distribution from data, or determine adequacy.

Distribution tails: the same moments, different worst days

An exact four-row witness with equal mean and variance

Input snapshot

Scale: 1
Distribution A: −1 and +1, each with probability $\frac12$
Distribution B: $-\frac12$ with probability $\frac45$ , and +2 with probability $\frac15$
Tail rule: Count support values whose absolute value is strictly greater than the threshold
Threshold multiple: 1.5

Deciding record

Deciding number: 20 percentage points
Equation: $P(\lvert A\rvert > k)=0,\qquad P(\lvert B\rvert > k)=\frac15,\qquad 1<k<2$ $\begin{gathered}P(\lvert A\rvert > k)=0, \\ P(\lvert B\rvert > k)=\frac15, \\ 1<k<2\end{gathered}$
Substitution: At $k = 1.5$ , neither ±1 lies outside. The value +2 does, carrying probability $\frac15 = 20\%$ .
Mean 0 and variance 1 stayed fixed: Mean 0 and variance 1 stayed fixed

Evidence table

Support and tail ledger: Support and tail ledger
Distribution Value Exact probability Outside threshold
A -1 $\frac{1}{2}$ No
A 1 $\frac{1}{2}$ No
B -0.5 $\frac{4}{5}$ No
B 2 $\frac{1}{5}$ Yes

Boundary and source

Boundary: Matching two moments alone does not identify a distribution or estimate a real tail.
Source: MIT Probability and Random Variables grounds the discrete probability calculation. NIST's normal distribution reference supports the repaired statement that mean and variance identify the member only after a suitable kind of distribution is fixed. No source wording is reproduced.
Model version: 1.0.0
Revision date: 31 August 2026

The moments match

The deciding record

Boundary and source

Open the moment check

The equal moments

Worked example

Related

Input snapshot

Deciding record

Evidence table

Boundary and source