I have a elevator probability problem but with a twist (instead of people exiting the elevator, it is the elevator stopping on each level). Require some help understanding and completing the probability questions.
Problem:
4 people go into the elevator of a 5-storey shophouse. Assume that each of them exits the building uniformly at random at any of the 5 levels and independently of each other. N is the random variable which is the total number of elevator stops.
- Describe the sample space for this random process.
Answer: {1,2,3,4,5}
Understanding: Sample space represents all the possible outcomes of the random event. Hence, the elevator can only stop at these 5 floors. Just like throwing a die and randomly getting a number, the sample space for that is {1,2,3,4,5,6}.
- Let $X_i$ be the random variable that equals to 1 if the elevator stops at floor i and 0, otherwise. Find the probability that the elevator stops at both floors $a$ and $b$ for $a$, $b$ $\in$ {1,2,3,4,5}. Find $EX_aX_b$.
Question: Do they mean the elevator stop at two consecutive floors or two different floors? Can a and b be the same, a smaller than b and vice versa?
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Prove the independence of $EX_1X_2$.
-
Calculate $EN^2$ (represented as ($X_1 + …+ X_5)^2 = \sum_{(i,j)} X_aX_b$ where the sum is over all ordered pairs (a,b) of numbers from {1,2,3,4,5} and the linearity of expectation. Find the variance of N.
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Determine the distribution of N. That is, determine the probabilities of events N = i. Compute $EN and EN^2$ directly by using the laws of expectation.
I got stuck from question 2 onwards with thinking of the possible ways that this can be done. It would be superb if someone can shed some light on them. Many thanks.
Best Answer
1)
Your answer only gives the set of values that can be taken by $N$ as outcomes. That is not enough in this situation.
You could go for $\Omega=\{0,1\}^5$ where e.g. $\omega=(1,0,0,1,0)$ is the outcome that the elevator stops on the floors $1$ and $4$.
Then for $i\in\{1,2,3,4,5\}$ we prescribe random variable $X_i:\Omega\to\mathbb R$ by $\omega=(\omega_1,\omega_2,\omega_3,\omega_4,\omega_5,)\mapsto\omega_i$.
Then the event "the elevator stops at floor $i$" can be recognized as the event $\{X_i=1\}$.
2)
I would rather avoid $a,b$ and choose for $i,j$ instead. Note that $X_iX_j=1$ if and only if the elevator stops at floor $i$ and stops at floor $j$. This with $\mathbb EX_iX_j=P(X_iX_j=1)$ because $X_iX_j$ only takes values in $\{0,1\}$.
If $i=j$ then $X_iX_j=X_i^2=X_i$ so in that case we must find $P(X_i=1)$.
This can be done by:$$P(X_i=1)=1-P(X_i=0)=1-\left(\frac45\right)^4$$
Observe here that $\{X_i=0\}$ is the event that all passengers exit on a floor different from $i$.
If $i\neq j$ then we go for calculation:$$P\left(X_{i}X_{j}=1\right)=P\left(X_{i}=1\wedge X_{j}=1\right)=1-P\left(X_{i}=0\vee X_{j}=0\right)=$$$$1-P\left(X_{i}=0\right)-P\left(X_{j}=0\right)+P\left(X_{i}=0\wedge X_{j}=0\right)=1-2\cdot\left(\frac{4}{5}\right)^{4}+\left(\frac{3}{5}\right)^{4}$$
3)
I don't know what is meant by "prove the independence of $\mathbb EX_1X_2$" since $\mathbb EX_1X_2$ is a real number. They might mean something like: "check that $\mathbb EX_iX_j$ is not depending on $(i,j)$ if $i\neq j$.
4)
$N^2=\sum_{i=1}^5\sum_{j=1}^5X_iX_j$ and this equality enables you to find an expression for $\mathbb EN^2$ by means of linearity of expectation and symmetry.
5)
Finding $P(N=i)$ for $i=1,2,3,4,5$ might turn out to be the hardest part of the question, but something must be left for you.
If you don't manage to do it yourself then please let me know.
edit (concerning determination of distribution of $N$).
Number the floors with $1,2,3,4,5$ using index variable $j$ for them.
Number the persons with $1,2,3,4$ using index variable $i$ for them.
Let $Y_{i,j}=1$ if person $i$ leaves the elevator at floor $j$ and let $Y_{i,j}=0$ otherwise.
Then $Y_{j}:=\sum_{i=1}^{4}Y_{i,j}$ equalizes the number of persons that leave the elevator at floor $j$.
Here $Y_{1,j},Y_{2,j},Y_{3,j},Y_{4,j}$ are iid and have Bernoulli distribution with parameter $\frac{1}{5}$.
Consequently $Y_{j}$ has binomial distribution with parameters $n=4$ and $p=\frac{1}{5}$.
Next to that we have: $$\sum_{j=1}^{5}Y_{j}=4\tag1$$
$Y:=\left(Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right)$ is a random vector where $Y_{j}$ denotes the number of persons that leave the elevator at floor $j$.
$Y$ has $5$-nomial distribution with parameters $n=4$ and probabilities $\left(p_{1},p_{2},p_{3},p_{4},p_{5}\right)=\left(\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5}\right)$.
This means that: $$P\left(Y=\left(r_{1},r_{2},r_{3},r_{4},r_{5}\right)\right)=5^{-4}\frac{4!}{r_{1}!r_{2}!r_{3}!r_{4}!r_{5}!}\tag2$$ for nonnegative integers $r_{1},\dots,r_{5}$ that satisfy $r_{1}+r_{2}+r_{3}+r_{4}+r_{5}=4$
We can define $N$ as the cardinality of the random set $S=\left\{ j\in\left\{ 1,2,3,4,5\right\} \mid Y_{j}>0\right\} $.
Based on $(1)$ we find:
$N=1\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 4,0\right\} $
$N=2\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,0\right\} \text{ or }\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 3,1,0\right\} $
$N=3\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,1,0\right\} $
$N=4\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 1,0\right\} $
This enables you to find the distribution of $N$.
As an example let us find $P\left(N=3\right)$.
In order to get $\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,1,0\right\} $ first observe on base of $(2)$ that: $$P\left(Y_{1}=2,Y_{2}=1,Y_{3}=1\right)=5^{-4}\frac{4!}{2!1!1!0!0!}=12\cdot5^{-4}$$
Then realize that $Y=(2,1,1,0,0)$ is just one of $5\binom{4}{2}=30$ possibilities to arrive at $\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,1,0\right\} $ so that finally we arrive at: $$P\left(N=3\right)=30\cdot12\cdot5^{-4}=360\cdot5^{-4}$$