# Bounding $E[\|\Sigma^{-1/2}(X-\mu)\|_2^3]$ for 2-dimensional Bernoulli

Let $$X\in\{0,1\}^2$$ have mean $$\mu=\left[\begin{smallmatrix}p_1\\p_2\end{smallmatrix}\right]$$ and $$\Pr[X_1 = X_2 = 1] = p\le \min\{p_1,p_2\}$$. (Note we must have $$1-p_1-p_2+p\ge 0$$ for the distribution to be well defined.) We can then compute the covariance matrix $$\Sigma = E[(X-\mu)(X-\mu)^T] = \left[\begin{smallmatrix}p_1(1-p_1)&p-p_1p_2\\p-p_1p_2&p_2(1-p_2)\end{smallmatrix}\right]$$.

I would like to use the Berry Essen bound, and for that we need to upper bound the quantity $$\gamma=E[\|\Sigma^{-1/2}(X-\mu)\|_2^3]$$.

I believe one should be able to show $$\gamma \le C\left(\tfrac1{\sqrt{p_1(1-p_1)}}+\tfrac1{\sqrt{p_2(1-p_2)}}+\tfrac1{\sqrt{\min\{p_1,p_2\}-p}}\right) ,$$ for some universal constant $$C>0$$.

The symbolic computation of $$\Sigma^{-1/2}$$ is a bit unwieldy though, and so I wonder if there is some tricks I may use to arrive at this result more neatly?

Or if not, any proof would be appreciated.

Update: Using Pinelli's observation, we can compute \begin{align}\gamma |\Sigma|^{3/2} &= p[(1-p_1)(1-p_2)v]^{3/2} \\&+ (p_1-p)[(1-p_1)p_2(1-v)]^{3/2} \\&+ (p_2-p)[p_1(1-p_2)(1-v)]^{3/2} \\&+ (1-p_1-p_2+p)[p_1 p_2 v]^{3/2} , \end{align} where $$v = p_1+p_2-2p$$ and $$|\Sigma|=p_1p_2(1-v)-p^2$$.

It seems like I forgot a factor $$1/\sqrt{p}$$. In fact it seems that bounds of $$\gamma \le \frac{2} {\sqrt{p\,(\min\{p_1,p_2\}-p)}}$$ or $$\gamma \le \frac{1} {\sqrt{p}}+ \frac{2} {\sqrt{\min\{p_1,p_2\}-p}}$$ are sufficient and correct.

Your inequality does not hold in general:

You don't need to compute to compute $$\Sigma^{-1/2}$$, because $$\|\Sigma^{-1/2} x\|_2^2=x^T\Sigma^{-1}x$$ for all $$x\in\mathbb R^2$$. Using this simple observation with e.g. $$p=0$$, $$p_1=1/2$$, $$p_2=1/2-\epsilon$$, and $$\epsilon\downarrow0$$, we find $$\gamma =\frac{\left(1-\epsilon -\epsilon ^2\right)^{3/2}}{\sqrt{\epsilon }}+\frac{1}{2} (1+\epsilon )^{3/2}+\frac{\left(1-\epsilon +2 \epsilon ^2\right)^{3/2}}{2 \sqrt{1-2 \epsilon }}\to\infty,$$ whereas $$\frac1{\sqrt{p_1(1-p_1)}}+\frac1{\sqrt{p_2(1-p_2)}}+\frac1{\sqrt{\min(p_1,p_2)-p}}\to4+\sqrt2.$$

• Good point, and good approach. It looks like the term I need is $\frac{1}{\sqrt{(p_1+p_2-2p)(1-p_1-p_2+2p)}}$ rather than $\frac{1}{\sqrt{\min\{p_1,p_2\}-p}}$. What do you think about that version? Apr 19, 2020 at 10:34
• @ThomasDybdahlAhle : I think the new version requires a rather different approach. Therefore and because your current question has been answered, the new version should be posted as a separate question. Apr 19, 2020 at 13:40
• My conjecture was disproved, and you did point in a good direction, but I still don't have a useful upper bound, which was the main point of the question. Do you have a suggestion for how to state the new question without too much overlap with this one? Apr 19, 2020 at 19:07
• @ThomasDybdahlAhle : Your original question was fully answered. Therefore, I think any additional questions you may have should be posted separately. Apr 21, 2020 at 16:50

We have the simplex $$0\le p, 0\le p_1-p, 0\le p_2-p, 0\le 1-p_1-p_2+p$$. From this, we can deduce $$v\in[0,1]$$ and so we may bound

\begin{align} \gamma |\Sigma|^{3/2} &= p[(1-p_1)(1-p_2)v]^{3/2} \\&\quad+ (p_1-p)[(1-p_1)p_2(1-v)]^{3/2} \\&\quad+ (p_2-p)[p_1(1-p_2)(1-v)]^{3/2} \\&\quad+ (1-p_1-p_2+p)[p_1 p_2 v]^{3/2} \\&\le p[(1-p_1)(1-p_2)v] \\&\quad+ (p_1-p)[(1-p_1)p_2(1-v)] \\&\quad+ (p_2-p)[p_1(1-p_2)(1-v)] \\&\quad+ (1-p_1-p_2+p)[p_1 p_2 v] \\&=2|\Sigma| . \end{align} In other words $$\gamma\le2|\Sigma|^{-1/2}$$.

This bound is tight except at the four 0-faces/corners of the simplex.

In particular, the determinant is 0 exactly in the 1-faces of the simplex (see the picture), and $$|\Sigma|^{-1/2}$$ approaches those as $$1/\sqrt\epsilon$$, which $$\gamma$$ does too.

The bound still isn't quite tight though, as it approaches the 0-faces/corners as $$1/\epsilon$$, while $$\gamma$$ approaches them as $$1/\sqrt\epsilon$$. (E.g. if we let $$p=\epsilon, p_1=2\epsilon, p_2=3\epsilon$$.)