Bernoulli random variables choice of values

41 Views Asked by At

It seems that typically the standard is to define a bernoulli random variable $X$ as

$$ X = \begin{cases} 1 & \text{with probability } p \\ 0 & \text{with probability } 1 - p \end{cases} $$

My question is, why do we choose $1$ and $0$ instead of $2$ and $0$ or $1$ and $-1$ and any other number for that matter? Depending on what you choose, your expectation and variance will change.

1

There are 1 best solutions below

0
On

I agree with David's comment saying that it makes no difference. However, one might see the standard definition you gave as more natural because of some important functions of $X$ and the intuition associated with them.

Let's look at the expected value as an example: first, it should be clear that the expected value of the random variable $X$ you defined is $p$ and that this result is preety intuitive. Now, let's see how changing the values that $X$ assumes can be prejudicual. If you recall that $$E(AX + B) = AE(X) + B$$ and define $$X^\prime=\begin{cases}\frac{\pi}{2p}, \text{with probability p};\\ \frac{\pi}{2(1-p)}, \text{with probability } 1-p, \end{cases}$$

then, no matter the value of $p$ we will have $E(X^\prime)=\pi$, which does not give that much of an insight.

In essence the information you gain is the same, but the intuition behind it becomes clearer when you don't add extra unecessary stuff (noise) to it.