Hi everyone.
Today I'm going to talk about Bernoulli process practice
number one.
In this problem, you are visiting a rain forest.
But unfortunately you have run out of insect repellent.
As a result, the probability of you getting mosquito bites
is really high.
At each second, the probability that a mosquito
will land on your neck is 0.5.
If a mosquito lands on your neck, the probability that it
will bite you is 0.2.
And the probability that it will never bother you is 0.8.
All of this happens independently among all
mosquitoes.
For part A of the problem, we're interested in finding
the expected value of the time between successive mosquito
bites and the variance of the time between successive
mosquito bites.
From the problem statement we know that the probability
distributions of getting mosquito bites at different
times are identically distributed and independent.
Therefore, the mosquito bites occur as a Bernoulli process
with parameter p, where p represents the probability of
getting a mosquito bite at each second.
And p can be calculated as the probability that a mosquito
lands on your neck at each second multiplied by the
probability that a mosquito will bite you, given that it
has landed on your neck.
And this is equal to 0.5 times 0.2, which is equal to 0.1.
Next let us define x as the time between successive
mosquito bites.
Because of the memory-less property of the Bernoulli
process, which means the probability of getting
mosquito bites at different times are independent, x is
equivalent to the time until the next mosquito bite.
And x is a geometrical random variable whose
PMF is like the following.
For all x, let's say equal to 0, the probabilities
are equal to 0.
For x equal to 1, the probability that it takes 1
second to the next mosquito bite is simply equal to p.
And for x equal to 2, the probability that it takes 2
seconds until the next mosquito bite is equal to 1
minus p times p.
And for x equal to 3, the probability that it takes 3
seconds until the next mosquito bite is equal to 1
minus p to the power of 2 times p.
Similarly, for x equal to k, the probability that it takes
k seconds until the next mosquito bite is equal to 1
minus p to the power of k minus 1 times p.
Therefore the expected value of x is equal to 1 over p,
which is equal to 1 over 0.1, which is equal to 10.
And the variance of x is equal to 1 minus p over p squared,
which is equal to 1 minus 0.1 over 0.1 squared, which is
equal to 90.
For part B of the problem, we're considering
another type of bug.
Similar to the case as the mosquitoes, here at each
second the probability that a tick will land on your neck is
equal to 0.1.
And if a tick lands on your neck, the probability that it
will bite you is equal to 0.7.
And the probability that it will never bother
you is equal to 0.3.
And all this happens independently among all ticks
and all mosquitoes.
So similar to the case as part A, where mosquito bites occurs
as a Bernoulli process with parameter p equal to 0.1, here
the tick bites also across a Bernoulli process with
parameter q equal to 0.1 times 0.7, which is equal to 0.07.
And q is the probability of getting a tick
bite at each second.
Therefore, the bug bites occurs as a merged process
from the mosquito bites and the tick bites.
And let r represent the parameter for the bug bites.
So here r is equal to the probability of getting either
a mosquito bite or a tick bite.
And this is equivalent to 1 minus the probability of
getting no mosquito bite and no tick bite.
Because the mosquito bites and the tick bites happens
independently, therefore this can be written as 1 minus the
probability of no mosquito bites times the probability of
no tick bites at each second.
And this is equal to 1 minus p times 1 minus q, which is p
plus q minus pq, which is equal to 0.1 plus 0.7 minus
0.1 times 0.7.
Which is equal to 0.163.
Next, let us define y as the time between
successive bug bites.
So similar as x in part a, here y is a geometric
distribution with parameter r.
And therefore the expected value of y is equal to 1 over
r, which is equal to 1 over 0.163.
That is approximately 6.135.
And the variance of y is equal to 1 minus r over r squared,
which is equal to 1 minus 0.163 over 0.163 squared.
And this is approximately 31.503.
So this gives us the expected value of the time between
successive bug bites and the variance of the time between
successive bug bites.
And this concludes our two days practice
on Bernoulli process.