# If $|E(X)|< 1$ and $E(X^2)<1$, can we have $1 - E(X^2) = (1 - E(X))^2$?

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 $$0< \mu_1 < 1$$.

Background

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.

• Given that substantial answers have appeared, it would be better to post a new question rather than modifying this one. – whuber Dec 1 '19 at 16:53
• 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. – Vincent Granville Dec 1 '19 at 17:15
• My question stats.stackexchange.com/questions/438710/… has been updated accordingly and is now sound. – Vincent Granville Dec 2 '19 at 2:55
• At this point, both formulas for $\sigma_X^2$ (model $Y$ and $Z$) have been corrected and tested empirically. – 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 $$\beta = 1$$.
• 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 $\mbox{Var}(Z)$. – Vincent Granville Nov 30 '19 at 5:07
• 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. – Vincent Granville Nov 30 '19 at 5:36
• @VincentGranville Details follow from the fact that $E(X) = Var(X) = \lambda$ for a Poisson distribution. Setting $\lambda = 0.5$ does the trick. – knrumsey Nov 30 '19 at 5:48
• 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. – Vincent Granville Nov 30 '19 at 5:48