Of course $X=0$ works, but I am looking for a non-singular solution. I haven't made much progress to solve this problem. However, let $\mu_2 = E(X^2)$ and $\mu_1 = E(X)$. For the equality to hold, we must have $\mu_2 = \mu_1 (2-\mu_1) > 0$. This is clearly impossible if $\mu_1 < 0$, thus we can focus on the case $0< \mu_1 < 1$.


Let's $X_1,X_2,X_3$ and so on be i.i.d. with same distribution as $X$. Let's define the following infinite sums: $$Z = X_1 + X_1 X_2 + X_1 X_2 X_3 +\cdots \\ Y=X_1 + X_2 X_3 + X_4 X_5 X_6 +\cdots$$ We have (see here why):

$$ \mbox{Var}(Z) = \frac{\mbox{Var}(X)}{(1-\mu_1)^2(1-\mu_2)} , \mbox{ Var}(Y)=\frac{\mbox{Var}(X)}{(1-\mu_1^2)(1-\mu_2)}$$

Let's use the symbols $\mu$ and $\sigma^2$ to denote the expectation and variance of the infinite sum, regardless as to whether it comes from model $Z$, or from model $Y$. To test whether a data fits the model $Z$ or $Y$, the statistic of the test is

$$T = \sigma^2\cdot\frac{(1-\mu_2)(1-\mu_1^2)}{\mu_2-\mu_1^2}$$ Here $\sigma^2$ is the empirical variance computed on the observations modeled by the infinite series ($Z$ or $Y$ depending on the model). $T$ is expected to be equal to $1$ if the data matches model $Y$. But both models result in the same $T$ only if $(1-\mu_1^2) = (1-\mu_1)^2$. Note that $\mu_X = \mu_1$ and $\sigma_X^2$ are easy to estimate, using some formulas, for instance $\mu_X = \mu_1 = \mu/(1+\mu)$, valid for both models. Also:

  • For model $Z$:

$$\sigma_X^2 = \frac{(1-\mu_1)^2(1-\mu_1^2)\sigma^2}{1+\sigma^2(1-\mu_1)^2} $$

  • For model $Y$:

$$\sigma_X^2 = \frac{(1-\mu_1^2)^2\sigma^2}{1+\sigma^2(1-\mu_1^2)}$$

Note: If correct, it would imply that $\mu > -\frac{1}{2}$ in all cases where convergence (for the infinite sum) occurs, whether you use model $Z$ or $Y$. Also, if $\mu_X = 0$ then $\mu =0$ (the converse is also true) and $\sigma_X^2 = \sigma^2/(1+\sigma^2)$ regardless of the model.

I realize that I posted the wrong question, due to a typo when copy/pasting a formula. It should have been "can we have $1-E^2(X) =(1-E(X))^2$ which has the obvious answer "yes only if $E(X) = 0$" (since the case $E(X) =1$ must be excluded.)The issue is still the same, that is, getting a statistical test that can discriminate between model $Y$ and model $Z$, and the answers posted by @knRumsey and @Henry to my question are correct, it's just that I posted the wrong question. Not sure how to best handle this. It definitely makes my problem easier, but I need somehow to update my question.

  • $\begingroup$ Given that substantial answers have appeared, it would be better to post a new question rather than modifying this one. $\endgroup$ – whuber Dec 1 '19 at 16:53
  • $\begingroup$ Thanks @Whuber. Wondering if it is possible not put the whole Note in my question in bold font, though I understand you want to emphasize that there is something wrong in my question. Also, for the same reasons, my new question (stats.stackexchange.com/questions/438710/…) has the same issue. And the formulas for $\sigma_X^2$ for models $Y$ and $Z$ took a while to be found and are valuable, though the one for model $Y$ must be updated (I am working on it.) The correction is backed by empirical evidence. $\endgroup$ – Vincent Granville Dec 1 '19 at 17:15
  • $\begingroup$ My question stats.stackexchange.com/questions/438710/… has been updated accordingly and is now sound. $\endgroup$ – Vincent Granville Dec 2 '19 at 2:55
  • $\begingroup$ At this point, both formulas for $\sigma_X^2$ (model $Y$ and $Z$) have been corrected and tested empirically. $\endgroup$ – Vincent Granville Dec 2 '19 at 15:46

$$1 - E(X^2) = (1 - E(X))^2$$ is equivalent to $$Var(X) = 2E(X)(1 - E(X))$$ and any distribution with this will satisfy your condition. You need $0 \le E(X) \le 1$ so the variance will be non-negative, and strict inequalities for the variance to be positive.

Simple examples include knrumsey's $E(X)=\frac12$, $Var(X)=\frac12$. Another is $E(X)=\frac13$, $Var(X)=\frac49$.

For actual distributions, you could choose any $k$ with $0 < k <1$ and then have examples such as

  • A normal distribution $N(k, 2k(1-k))$ such as $N\left(\frac13,\frac49\right)$
  • A gamma distribution with $\alpha= \frac{k}{2(1-k)}$, $\beta=\frac{1}{2(1-k)}$ such as $\alpha= \frac{1}{4}$, $\beta=\frac{3}{4}$
  • A two-point distribution with $X= k \pm \sqrt{2k(1-k)}$ each with probability $\frac12$, such as $-\frac13$ and $+1$ with equal probability

and there are many more


This holds for any distribution with $E(X) = Var(X) = \frac{1}{2}$

First note that $E(X^2) = E(X)^2 + Var(X)$, so that your desired equality can be rewritten as $$\mu_1^2 + \sigma_X^2 = \mu_1(2-\mu_1)$$ Now set $\mu_X \stackrel{\cdot}{=} \mu_1 = \sigma_X^2$ and this becomes $\mu_X^2 + \mu_X = \mu_X(2-\mu_X)$ which has a solution for $\mu_X=\frac{1}{2}$.

It is easy to see that this satisfies all of the desired properties.

$$|E(X)| = \frac{1}{2} \quad\quad E(X^2) = \frac{3}{4} < 1$$ $$1-E(X^2) = \frac{1}{4} \quad\quad ((1-E(X))^2 = \frac{1}{4}$$

A few examples

  • $X$ is Normal with $\mu=0.5$ and $\sigma^2 = 0.5$.
  • $X$ is Poisson with $\lambda = 0.5$.
  • $X$ is Laplace with $\mu=0.5$ and $b = 0.5$.
  • $X$ is Gamma with $\alpha = 0.5$ and $\beta = 1$.
  • $\begingroup$ Thank you. Please could you edit your answer and replace $\sigma^2$ by $\sigma_X^2$? The reason is that I updated my question to use a simplied notation, and I am using $\sigma^2$ either for $\mbox{Var}(Y)$ or $\mbox{Var}(Z)$. $\endgroup$ – Vincent Granville Nov 30 '19 at 5:07
  • $\begingroup$ You mentioned a Poisson distribution in an earlier version of your answer, but I don't remember the details. Would like to see the details again, thanks. $\endgroup$ – Vincent Granville Nov 30 '19 at 5:36
  • 1
    $\begingroup$ @VincentGranville Details follow from the fact that $E(X) = Var(X) = \lambda$ for a Poisson distribution. Setting $\lambda = 0.5$ does the trick. $\endgroup$ – knrumsey Nov 30 '19 at 5:48
  • 1
    $\begingroup$ Thanks for the update. It looks like I need to find a better statistic (than $T$) for my test, since it leads to some ambiguity. $\endgroup$ – Vincent Granville Nov 30 '19 at 5:48

Your Answer

