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Let $X$ and $Y$ are two independent binomial random variables where $X\sim B(K, q), Y\sim B(K, p)$. I am wondering how to compute or estimate the following expectations: $$ \ \mathbb{E}[|pX-qY|]\ \ \text{and}\ \ \mathbb{E}[\text{ReLU}(pX-qY)].$$

where ReLU($x$) = max($x, 0$) Furthermore, what does the distribution of $pX-qY$ look like?

Update: Actually I'm more interested in $\mathbb{E}[pX\mathbb{1}(pX-qY>0)]\ \text{and}\ \mathbb{E}[qY\mathbb{1}(pX-qY>0)]$, where $\mathbb{1}$ is the indicator function. We know that $\mathbb{E}[pX\mathbb{1}(pX-qY>0)]-\mathbb{E}[qY\mathbb{1}(pX-qY>0)]=\mathbb{E}[\text{ReLU}(pX-qY)]$ for which @PedroSebe has presented a nice idea below.

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  • $\begingroup$ Cross-posted at math.stackexchange.com/q/3580440/321264. $\endgroup$ Mar 14 '20 at 7:46
  • $\begingroup$ There doesn't seem to be a closed form. $\endgroup$
    – gunes
    Mar 14 '20 at 9:05
  • $\begingroup$ @gunes Thanks! I have modified the question a bit. $\endgroup$
    – luw
    Mar 14 '20 at 14:47
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Although there is no closed form, if $K$ is big enough we can use Central Limit Theorem to approximate $X\sim\mathcal{N}(Kq,Kq(1-q))$ and $Y\sim\mathcal{N}(Kp,Kp(1-p))$. It follows that $pX-qY$ has asymptotically normal distribution with mean zero and variance given by:

$$\sigma^2=p^2\cdot Kq(1-q)+q^2\cdot Kp(1-p) \\=Kpq\left[p(1-q)+q(1-p)\right]\\=Kpq(p+q-2pq)$$

Then, using a property of the normal distribution, we get: $$\mathbb{E}[|pX-qY|]=\sqrt{\frac{2}{\pi}}\sigma=\sqrt{\frac{2Kpq}{\pi}(p+q-2pq)}$$

Now, for the ReLU, let us denote $Z=pX-qY$, for clarity. Then:

$$\mathbb{E}[\text{ReLU}(Z)]=\mathbb{E}[Z|Z>0]\cdot\mathbb{P}[Z>0]$$

Since $\text{ReLU}(Z)=0$ if $Z<0$ and $=Z$ otherwise. Then, using the fact that our distribution is symmetric around zero, we have:

$$\mathbb{E}[\text{ReLU}(Z)]=\frac{1}{2}\mathbb{E}[|pX-qY|]=\sqrt{\frac{Kpq}{2\pi}(p+q-2pq)}$$

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  • $\begingroup$ Thank you for your answer! Actually I'm more interested in $\mathbb{E}[pX\mathbb{1}(pX-qY>0)]\ \text{and}\ \mathbb{E}[qY\mathbb{1}(pX-qY>0)]$. (I have updated my question a bit.) Do you have any advice for this? $\endgroup$
    – luw
    Mar 15 '20 at 6:31

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