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