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For simplicity, I'll focus on the bivariate case. Let $(X_1,X_2)$ be a random vector that obeys bivariate Bernoulli. $X_i$ takes either zero or one. The associated pdf can be written as $$p(x_1,x_2)=p_{11}^{x_1x_2}p_{10}^{x_1(1-x_2)}p_{01}^{(1-x_1)x_2}p_{00}^{(1-x_1)(1-x_2)}.$$

Now, consider a categorical random variable $Y$ that takes four values $\{11,10,01,00\}$ with probability $\{p_{11},p_{10},p_{01},p_{00}\}.$

The associated pdf can be written as

$$p(y)=p_{11}^{[y=11]}p_{10}^{[y=10]}p_{01}^{[y=01]}p_{00}^{[y=00]},$$ where $[y=z]=1$ if and only if $y=z$.

So, it looks like any bivariate Bernoulli random vector can be represented using a categorical random variable.

However, if we think about the following multivariate Bernoulli random vector $Z$, the categorical distribution can also be represented using a multivariate Bernoulli.

Let $Z=(Z_1,Z_2,Z_3,Z_4).$ Each $Z_i$ is a Bernoulli variable that takes either zero or one. Z differs from the general multivariate Bernoulli in that only one of the four variables can take value one.

The pdf of this random vector can be written as

$$p(z_1,z_2,z_3,z_4)=p_{1000}^{z_1(1-z_2)(1-z_3)(1-z_4)}p_{0100}^{(1-z_1)z_2(1-z_3)(1-z_4)}p_{0010}^{(1-z_1)(1-z_2)z_3(1-z_4)}p_{0001}^{(1-z_1)(1-z_2)(1-z_3)z_4}.$$

Now, we have a multivariate Bernoulli random vector that represents the categorical variable in the above.

My question is what is the relationship between the two random variable/vector and their associated distributions?

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  • $\begingroup$ What kind of relationships are you looking for? I could for instance say the correlation between $X_{1}$ and $Z_{1}$ is $p_{11}/(p_{11} + p_{10})$. I'm not sure how much more can be said? $\endgroup$ Jul 22, 2020 at 1:46
  • $\begingroup$ @SherwinLott What I want to know is if I have a multivariate Bernoulli random vector, can I take it as a categorical random variable with the appropriate conversion of the sample space like what I did in the bivariate example? And similarly, if I have a categorical random variable, can I convert it into a multivariate Bernoulli with a restriction that only one of the variable takes value one? $\endgroup$
    – Andeanlll
    Jul 22, 2020 at 1:51
  • $\begingroup$ Yes, all of your conversions seem perfectly reasonable. You could write $p(z_1,z_2,z_3,z_4)$ out as the multiplication of all 16 probabilities to make it look like a multivariate binomial, so that you don't have to explicitly constrain $Z_{1} + Z_{2} + Z_{3} + Z_{4} = 1$. All the other 12 probabilities are zero, and they would evaluate to $1$ when raised to the zeroth power. Was that the hangup? You are concerned about explicitly constraining the $Z$'s to add up to $1$? This way it would be implicit and you'd be writing out the multivariate binomial pdf canonically. $\endgroup$ Jul 22, 2020 at 2:01
  • $\begingroup$ @SherwinLott I'm curious about why we have two different random variables when one can be perfectly described by another.. I thought there's some point that I was missing. Maybe an example of a multivariate Bernoulli that cannot be converted into a categorical variable or a categorical variable that can't be represented by a multivariate Bernoulli. $\endgroup$
    – Andeanlll
    Jul 22, 2020 at 2:52
  • $\begingroup$ I understand what you're saying, so here's why it's not surprising to me. The more random variables you have the more scenarios you can represent. You can represent any bivariate distribution with the multivariate here. What would be really surprising though is if you could represent any multivariate with a bivariate, which obviously cannot be done. It's not surprising that adding information/variables doesn't stop you from encoding the same information as before. $\endgroup$ Jul 22, 2020 at 3:12

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Focusing on the $n=2$ case

Let me introduce the following probability mass function: \begin{align*} p(y_1, y_2) = \pi_1^{y_1}(1-\pi_1)^{1-y_1}\pi_2^{y_2}(1-\pi_2)^{1-y_2}\left(1 + \rho \frac{(y_1 - \pi_1)(y_2 - \pi_2)}{\sqrt{\pi_1(1-\pi_1)\pi_2(1-\pi_2)}}\right) \end{align*} known as Bahadur's model. You can indeed verify that \begin{align*} \sum_{(y_1, y_2) \in \{0, 1\}^2} p(y_1, y_2) &= 1 \\ \text{Corr}(Y_1, Y_2) &= \rho \end{align*} There is a bijection between $(p_{11}, p_{10}, p_{01}, p_{00})$ and $(\pi_1, \pi_2, \rho)$ through the relations \begin{align*} p_{11} &= \pi_1\pi_2\left(1 + \rho\frac{(1-\pi_1)(1-\pi_2)}{\sqrt{\pi_1(1-\pi_1)\pi_2(1-\pi_2)}}\right) \\ p_{10} &= \pi_1(1-\pi_2)\left(1 - \rho\frac{(1-\pi_1)\pi_2}{\sqrt{\pi_1(1-\pi_1)\pi_2(1-\pi_2)}}\right) \\ p_{01} &= (1-\pi_1)\pi_2\left(1 - \rho\frac{\pi(1-\pi_2)}{\sqrt{\pi_1(1-\pi_1)\pi_2(1-\pi_2)}}\right) \\ p_{00} &= (1-\pi_1)(1-\pi_2)\left(1 + \rho\frac{\pi_1\pi_2}{\sqrt{\pi_1(1-\pi_1)\pi_2(1-\pi_2)}}\right) \end{align*} so Bahadur's model is just a parametrization of the bivariate binary model. Now let $\rho = -1$ and $\pi_1 = 1 - \pi_2 = \pi$. This gives \begin{align*} p_{11} &= 0 \\ p_{10} &= \pi\\ p_{00} &= 0 \\ p_{01} &= 1-\pi \end{align*} So, the two-category categorical model is just a special case of Bahadur's model when the correlation is $\rho = -1$. This makes sense; a categorical random variable is basically a multivariate binary with hugely negative correlations among the entries to force only one selected category. We use this to generalize the result.

Generalizing the result

Bahadur's model can be expanded to multivariate binary random variables $p(y_1, \cdots, y_n)$ with the representation \begin{align*} p(y_1, \cdots, y_n) = \left[\prod_{i=1}^n\pi_i^{y_i}(1-\pi_i)^{1-y_i}\right]\left(1 + \sum_{k=2}^{n}\rho_k\text{Sym}_k(\mathbf{r}_n)\right) \end{align*} where \begin{align*} r_i &= \frac{y_i - \pi_i}{\sqrt{\pi_i(1-\pi_i)}} \\ \mathbf{r}_n &= (r_1, \cdots, r_n) \\ \text{Sym}_k(\mathbf{r}_n) &= \sum_{i_1<\cdots<i_k}r_{i_1}\cdots r_{i_k} \end{align*} I'm not entirely sure what choice of the parameters can lead to a genuine categorical random variable (will think about this and post if I have a positive result), but this is a starting place.

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