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In this problem, we're given a
random variable X which has a

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uniform distribution in the
interval negative 1 to 1.

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In other words, if we were to
draw out the PDF of X, we see

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that in the interval negative
1 to 1, it has value 1/2.

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Now we're given a sequence
random variables X1, X2, and

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so on, where each Xi has the
same distribution as X and

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different Xi's are
independent.

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For part a, we would like to
know if the sequence Xi

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converges to some number--

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let's call it c--

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in probability as i
goes to infinity--

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whether this is true.

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Let's first recall the
definition of convergence in

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probability.

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If this does happen, then by
definition, we'll have that

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for every epsilon greater than
0, the probability Xi minus c

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greater equal to epsilon, this
quantity will go to 0 in the

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limit of i going to infinity.

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In other words, with very high
probability, we will find Xi

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to be very concentrated around
the number c if this were to

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be the PDF of Xi.

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Now, can this be true?

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Well, we know that each Xi is
simply a uniform distribution

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over negative 1 to 1.

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It doesn't really change
as we increase i.

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So intuitively, the
concentration around any

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number c is not going
to happen.

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So we should not expect a
convergence in probability in

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this sense.

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For part b, we would like to
know whether the sequence Yi,

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defined as Xi divided by i,
converges to anything in

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probability.

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Well, by just looking at the
shape of Yi, we know that

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since the absolute value of Xi
is less than 1, then we expect

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the absolute value of
Yi is less than 1/i.

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So eventually, Yi gets
very close to 0

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as i goes to infinity.

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So it's safe to bet that maybe
Yi will converge to 0 in

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probability.

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Let's see if this is
indeed the case.

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The probability of Yi minus 0
greater equal to epsilon is

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equal to the probability of Yi
absolute value greater equal

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to epsilon.

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Now, previously we know that the
absolute value of Yi is at

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most 1/i by the definition
of Yi.

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And hence the probability right
here is upper bounded by

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the probability of 1i greater
equal to epsilon.

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Notice in this expression,
there is nothing random.

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i is simply a number.

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Hence this is either 1 if i is
less equal to 1/epsilon, or 0

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if i is greater than
1/epsilon.

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Now, this tells us, as long as
i is great enough-- it's big

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enough compared to epsilon--

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we know that this quantity
here is [INAUDIBLE]

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0.

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And that tells us in the limit
of i goes to infinity

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probability of Yi deviating
from 0 by more than

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epsilon goes to 0.

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And that shows that indeed, Yi
converges to 0 in probability

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because the expression right
here, this limit, holds for

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all epsilon.

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Now, in the last part of the
problem, we are looking at a

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sequence Zi defined by Xi raised
to the i-th power.

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Again, since we know Xi is some
number between negative 1

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and 1, this number raised to the
i-th power is likely to be

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very small.

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And likely to be small in the
sense that it will have

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absolute value close to 0.

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So a safe guess will be the
sequence Zi converges to 0 as

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well as i goes to infinity.

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How do we prove this formally?

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We'll start again with a
probability that Zi stays away

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from 0 by more than epsilon
and see how that evolves.

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And this is equal to the
probability that Xi raised to

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the i-th power greater
equal to epsilon.

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Or again, we can write this by
taking out the absolute value

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that Xi is less equal to
negative epsilon raised to the

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1 over i-th power or Xi greater
equal to epsilon 1

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over i-th power.

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So here, we'll divide into two
cases, depending on the value

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of epsilon.

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In the first case, epsilon
is greater than 1.

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Well, if that's the case, then
we know epsilon raised to some

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positive power is still
greater than 1.

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But again, Xi cannot have any
positive density be on the

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interval negative 1 or 1.

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And hence we know the
probability above, which is Xi

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less than some number smaller
than negative 1 or greater

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than some number bigger
than 1 is 0.

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So that case is handled.

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Now let's look at a case where
epsilon is less than 1,

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greater than 0.

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So in this case, epsilon to the
1/i will be less than 1.

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And it's not that difficult
to check that since Xi has

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uniform density between negative
1 and 1 of magnitude

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1/2, then the probability here
was simply 2 times 1/2 times

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the distance between epsilon
to the 1 over

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i-th power and 1.

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So in order to prove this
quantity converge to 0, we

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simply have to justify why
does epsilon to the 1/i

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converge to 1 as i
goes to infinity.

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For that, we'll recall
the properties

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of exponential functions.

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In particular, if a is a
positive number and x is its

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exponent, if we were to take the
limit as x goes to 0 and

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look at the value of a to the
power of x, we see that

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this goes to 1.

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So in this case, we'll let a be
equal to epsilon and x be

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equal to 1/i.

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As we can see that as i goes to
infinity, the value of x,

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which is 1/i, does go to 0.

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And therefore, in the limit i
going to infinity, the value

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of epsilon to the 1 over
i-th power goes to 1.

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And that shows if we plug this
limit into the expression

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right here that indeed, the term
right here goes to 0 as i

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goes to infinity.

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And all in all, this implies the
probability of Zi minus 0

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absolute value greater equal to
epsilon in the limit of i

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going to infinity converges to
0 for all positive epsilon.

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And that completes our proof
that indeed, Zi converges to 0

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in probability.

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