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Suppose that we have two random variables consisting of an indicator function like below: \begin{align*} V_1=1[c_1\geq u_1] \\ V_2=1[c_2\geq u_2] \end{align*} where $c_1$ and $c_2$ are constants, and $u_1$ and $u_2$ follow the standard normal variables.

Thus, $\Pr[V_1=1]=\Phi(c_1)$ and $\Pr[V_2=1]=\Phi(c_2)$, and as we can see, $V_1$ and $V_2$ follow Bernoulli distributions with the probability $\Phi(c_1)$ and $\Phi(c_2)$, respectively.

Q 1. Here, I would like to know the distribution of $V_1+V_2$ when $u_1$ and $u_2$ are not independent. (If they are independent, then the distribution may be binomial)

Q 2. Additionally, I am also wondering whether $V_1+V_2$ can be expressed using another indicator function like $V_1+V_2=1[c_2\geq u_3]$.

I guess, it is impossible because that it is possible implies that the sum of two Bernoulli follows a Bernoulli distribution.

If my conjecture is correct, then is it possible to express $V_1+V_2$ using only one random variable $u_3$? (This question is possibly equivalent to Q 1)

I have been striving to solve the questions above, but so far no progress has been made.

Thank you.

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Q1: In general, dependency between $V_1$ and $V_2$ can be dealt with by writing the joint probabilities of the outcomes of these variables:

$$\begin{matrix} \theta_{00} \equiv \mathbb{P}(V_1=0, V_2=0) & & & \theta_{01} \equiv \mathbb{P}(V_1=0, V_2=1), \\[6pt] \theta_{10} \equiv \mathbb{P}(V_1=1, V_2=0) & & & \theta_{11} \equiv \mathbb{P}(V_1=1, V_2=1). \\[6pt] \end{matrix}$$

The distribution of the sum $V_1+V_2$ is then given by:

$$\begin{align} \mathbb{P}(V_1+V_2 = 0) &= \theta_{00}, \\[6pt] \mathbb{P}(V_1+V_2 = 1) &= \theta_{01} + \theta_{10}, \\[6pt] \mathbb{P}(V_1+V_2 = 2) &= \theta_{11}. \\[6pt] \end{align}$$

In the case where $V_1$ and $V_2$ depend on underlying random variables, the probabilities $\theta_{00},\theta_{01},\theta_{10},\theta_{11}$ can be written as functions of the underlying joint distribution of those random variables, and you would then obtain the required form by substitution of the expressions for those probabilities.


Q2: No, this is not possible. As you correctly point out, this form would give $\mathbb{P}(V_1+V_2 = 2)=0$, which does not capture the distribution correctly.

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  • $\begingroup$ Thank you so much for your clear answer! $\endgroup$ Aug 31, 2023 at 1:58

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