3
$\begingroup$

Given $n$ people, each has probability $v$ of having a virus. Of those with the virus, they are hospitalized with probability $p$. Independently of having the virus, any of the $n$ people may be hospitalized for a different reason with probability $a$.

Let $X$ be a random variable that denotes the number of the $n$ people who are hospitalized, and $Y$ be a random variable that denotes the number of people in the hospital that have the virus.

Find the distributions of X and Y.

I think this situation reflects a certain distribution. I was thinking Bernoulli, but I can't seem to think of a way to represent both of the random variables. Can anyone help with this please?

$\endgroup$
0
$\begingroup$

There are different ways to think about this problem. One way is to think in terms of Bernoulli variables. Consider just a single person. Let $U = 1$ if he has the virus and zero otherwise. Then $U \sim \text{Bernoulli}(v)$. Let $V \sim \text{Bernoulli}(p)$ independent of $U$, modeling the chance of hospitalization due to the virus. Then $UV$ is a Bernoulli variable which is one if the person is hospitalized due to the virus.

Now, let $W \sim \text{Bernoulli}(a)$ independent of $U$ and $V$. Here, $W$ models the chance of hospitalization independent of the virus. Then, you can check that $UV \vee W$ is the indicator of the overall hospitalization. (Here $\vee$ is the logical OR operator, or in other words the maximum of its two arguments.)

(There might be easier ways of dealing with the problem.)

EDIT: It might be easier to look at the indicator of not being hospitalized which is $1 - (UV \vee W) = (1-UV) (1-W)$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$X$ and $Y$ have a binomial distribution. As @passerby51 mentions, the event that an individual is hospitalized or hospitalized with the virus is distributed Bernoulli. The sum of Bernoulli distributions is binomial.

Assuming the ways of being hospitalized are mutually exclusive, the probability any one person is hospitalized is $pv+a$ and the probability they are hospitalized with the virus is $pv$.

The probability $X$ are hospitalized is given by the binomial distribution $B(n,pv+a)$. The $pmf$ is: $$P(X=x)= {n\choose x}\left(pv+a\right)^x\left(1-pv-a\right)^{n-x}$$

Similarly, $Y$ is distributed binomial $B(n,pv)$. The $pmf$ is: $$P(X=x)= {n\choose x}\left(pv\right)^x\left(1-pv\right)^{n-x}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $pv + a$ can be bigger than one, so it is not the right probability. $\endgroup$ – passerby51 May 9 '13 at 11:24
  • $\begingroup$ This is embedded in my assumption that the two events are mutually exclusive. If they are not, we ought to subtract that probability that a person is hospitalized due to the virus and due to some other reason simultaneously. $\endgroup$ – CommonerG May 9 '13 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.