Are there any continuous probability distributions whose random variables have variances that do not change as you raise them to higher and higher integer powers? If there are, what are they?

I've done some simulations, and a few things are clear. For a uniform distribution with min = 0 and max = 1, the variances of the random variables converge to 0 as higher and higher positive integer powers are used (I know that this is expected since raising numbers to positive integer powers that are between 0 and 1 makes them smaller). For a Cauchy distribution with location = 0 and scale = 1, variances diverge to Inf (I know that this is expected since Cauchy distributions have variances of infinity). It's not clear to me if the variances of a normal distribution with mean = 0 and sd = 1 converge or diverge.

It'd be great if I could find a probability distribution (or multiple probability distributions) whose random variables have variances that remain constant as higher and higher integer powers are used. To be clear, I'm not looking for variances that converge - I'm looking for variances that remain constant. I'm looking for a theoretical basis and not just an empirical one.

Thanks! My R code is below for reference.

# Constants
Highest_Integer_Power_to_Investigate <- 10
Number_or_Observations_per_Sample <- 1000
Number_of_Iterations <- 1000

# Data Generation
Uniform_Distribution_Values <- lapply(seq_len(Number_of_Iterations), function (x) {
Normal_Distribution_Values <- lapply(seq_len(Number_of_Iterations), function (x) {
Cauchy_Distribution_Values <- lapply(seq_len(Number_of_Iterations), function (x) {
Values <- list(Uniform_Distribution = Uniform_Distribution_Values, Normal_Distribution = 
Normal_Distribution_Values, Cauchy_Distribution = Cauchy_Distribution_Values)

# Calculations
Output <- lapply(Values, function (x) {
  sapply(x, function (y) {
    sapply(seq_len(Highest_Integer_Power_to_Investigate), function (z) {
      var(y ^ z)

# Formatting
Variances <- as.data.frame(lapply(Output, rowMeans))
Variances$Power <- seq_len(Highest_Integer_Power_to_Investigate)
Variances <- Variances[, c(which(colnames(Variances) == 'Power'), which(colnames(Variances) != 'Power'))]

# Output
#    Power Uniform_Distribution Normal_Distribution Cauchy_Distribution
# 1      1           0.08336001        1.003362e+00        2.004660e+06
# 2      2           0.08886279        2.013401e+00        1.954473e+18
# 3      3           0.08029366        1.515248e+01        2.209213e+30
# 4      4           0.07103040        9.744815e+01        2.626589e+42
# 5      5           0.06304486        9.661652e+02        3.208136e+54
# 6      6           0.05642795        1.046558e+04        3.973433e+66
# 7      7           0.05095627        1.386277e+05        4.955865e+78
# 8      8           0.04639468        2.018680e+06        6.202649e+90
# 9      9           0.04255073        3.260260e+07       7.776300e+102
# 10    10           0.03927592        5.601374e+08       9.757302e+114
  • 1
    $\begingroup$ Well, you could start with a discrete distribution on $\{0,1\}$. Good luck with finding anything else! $\endgroup$ Jun 21, 2023 at 15:54
  • $\begingroup$ Writing $\mu_{X^{2n}}=\sigma+\mu_{X^n}=\cdots=k\sigma+\mu_{X^m}$ ($n=2^km$, $m$ odd), we can specify the odd moments $\mu_{X^m}$, which will determine all moments, hence the mgf, hence a pdf via inversion formula. $\endgroup$
    – coiso
    Jun 21, 2023 at 16:07
  • $\begingroup$ @elemelons - I follow what you're saying but I'm not capable of doing it on my own. Can you expand a bit? $\endgroup$ Jun 21, 2023 at 17:28
  • $\begingroup$ Doing what on your own, precisely? Picking just the right values for odd moments so that we can explicitly invert the mgf to an elementary pdf? I dunno how to do that either. If you get software to compute inverse Fourier / Laplace transforms numerically you could pick values for odd moments and then approximate the pdf. $\endgroup$
    – coiso
    Jun 21, 2023 at 17:31
  • $\begingroup$ Could you clarify what you mean by raising a distribution to a power? Are you multiplying the random variable or the probability density function? $\endgroup$
    – Karl
    Jun 22, 2023 at 13:51

1 Answer 1


I believe that the only random variables that have the property that the variance of powers of remain constant are constant random variables, and discrete random variables taking values in $\{0,1\}$ (i.e. the only non--degenerate random variable such that $X^k = X$ for all $k$).

I'm not sure what your familiarity is with rigorous probability is, but the mathematical idea behind establishing this is the following. Consider for the moment positive random variables $X \ge 0$. Evidently constant random variables will satisfy $Var(X^k)=0$. For a non-degenerate random variable taking two values, say $a,b \in [0,\infty)$, $a<b$, let us use the notation $P(X=a) =p_1$, $P(X=b)=p_2$, $p_1+p_2=1$, $0 < p_1 ,p_2 <1$. Note here the last condition that both probabilities are strictly between 0 and 1 describes "non-degeneracy". We have

$$ Var(X^k) = E(X^{2k}) - [E(X^k)]^2. $$

We see that $$E(X^{2k}) = a^{2k}p_1 + b^{2k} p_2, \;\; E(X^{k})^2 = a^{2k}p_1^2 + a^kb^kp_1p_2+ b^{2k} p_2^2. $$ Evidently under these conditions, if $a,b \in (0,1)$, $Var(X^k) \to 0$ as $k\to \infty$. If $a,b\in \{0,1 \}$, $Var(X^k) = Var(X)$ is a constant, and if $b >1$, $Var(X^k) \to \infty$ as $k\to \infty$, since $p_2^2 < p_2$.

This calculation can be taken a little further. If $X_m$ is a random variable that takes finitely many values, $a_1 < ... < a_m $, so that $X_m = \sum_{i=1}^m a_i 1_{A_i}$,

$$E(X_m^{2k}) = \sum_{i=1}^m a_i^{2k}P(A_i) \sim a_m^{2k}P(A_m), \;\; E(X_m^{k})^2 = \left[\sum_{i=1}^m a_i^{k}P(A_i)\right]^2 \sim a_m P(A_m)^2 . $$

The same conclusions hold once again. Now if $X$ is a (positive) continuous random variable, it can be approximated by a random variable $X_m$ taking finitely many values (this is a standard result from measure--theoretic probability). If $X$ satisfies that $P(X\in [0,1]) =1$, this approximation can be made so that $a_m \in (0,1)$, and the same conclusion can be obtained that $Var(X^k) \to 0$. If instead $P(X > 1) > 0$, then $X_m$ may be taken so that $a_m >1$, and we will get that $Var(X^k) \to \infty$.

Since even powered random variables are positive, having considered positive random variables is enough to establish that powers of non--degenerate, continuous random variables will not have stable variances.


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