In this problem, we're given a random variable X which has a
uniform distribution in the interval negative 1 to 1.
In other words, if we were to draw out the PDF of X, we see
that in the interval negative 1 to 1, it has value 1/2.
Now we're given a sequence random variables X1, X2, and
so on, where each Xi has the same distribution as X and
different Xi's are independent.
For part a, we would like to know if the sequence Xi
converges to some number--
let's call it c--
in probability as i goes to infinity--
whether this is true.
Let's first recall the definition of convergence in
probability.
If this does happen, then by definition, we'll have that
for every epsilon greater than 0, the probability Xi minus c
greater equal to epsilon, this quantity will go to 0 in the
limit of i going to infinity.
In other words, with very high probability, we will find Xi
to be very concentrated around the number c if this were to
be the PDF of Xi.
Now, can this be true?
Well, we know that each Xi is simply a uniform distribution
over negative 1 to 1.
It doesn't really change as we increase i.
So intuitively, the concentration around any
number c is not going to happen.
So we should not expect a convergence in probability in
this sense.
For part b, we would like to know whether the sequence Yi,
defined as Xi divided by i, converges to anything in
probability.
Well, by just looking at the shape of Yi, we know that
since the absolute value of Xi is less than 1, then we expect
the absolute value of Yi is less than 1/i.
So eventually, Yi gets very close to 0
as i goes to infinity.
So it's safe to bet that maybe Yi will converge to 0 in
probability.
Let's see if this is indeed the case.
The probability of Yi minus 0 greater equal to epsilon is
equal to the probability of Yi absolute value greater equal
to epsilon.
Now, previously we know that the absolute value of Yi is at
most 1/i by the definition of Yi.
And hence the probability right here is upper bounded by
the probability of 1i greater equal to epsilon.
Notice in this expression, there is nothing random.
i is simply a number.
Hence this is either 1 if i is less equal to 1/epsilon, or 0
if i is greater than 1/epsilon.
Now, this tells us, as long as i is great enough-- it's big
enough compared to epsilon--
we know that this quantity here is [INAUDIBLE]
0.
And that tells us in the limit of i goes to infinity
probability of Yi deviating from 0 by more than
epsilon goes to 0.
And that shows that indeed, Yi converges to 0 in probability
because the expression right here, this limit, holds for
all epsilon.
Now, in the last part of the problem, we are looking at a
sequence Zi defined by Xi raised to the i-th power.
Again, since we know Xi is some number between negative 1
and 1, this number raised to the i-th power is likely to be
very small.
And likely to be small in the sense that it will have
absolute value close to 0.
So a safe guess will be the sequence Zi converges to 0 as
well as i goes to infinity.
How do we prove this formally?
We'll start again with a probability that Zi stays away
from 0 by more than epsilon and see how that evolves.
And this is equal to the probability that Xi raised to
the i-th power greater equal to epsilon.
Or again, we can write this by taking out the absolute value
that Xi is less equal to negative epsilon raised to the
1 over i-th power or Xi greater equal to epsilon 1
over i-th power.
So here, we'll divide into two cases, depending on the value
of epsilon.
In the first case, epsilon is greater than 1.
Well, if that's the case, then we know epsilon raised to some
positive power is still greater than 1.
But again, Xi cannot have any positive density be on the
interval negative 1 or 1.
And hence we know the probability above, which is Xi
less than some number smaller than negative 1 or greater
than some number bigger than 1 is 0.
So that case is handled.
Now let's look at a case where epsilon is less than 1,
greater than 0.
So in this case, epsilon to the 1/i will be less than 1.
And it's not that difficult to check that since Xi has
uniform density between negative 1 and 1 of magnitude
1/2, then the probability here was simply 2 times 1/2 times
the distance between epsilon to the 1 over
i-th power and 1.
So in order to prove this quantity converge to 0, we
simply have to justify why does epsilon to the 1/i
converge to 1 as i goes to infinity.
For that, we'll recall the properties
of exponential functions.
In particular, if a is a positive number and x is its
exponent, if we were to take the limit as x goes to 0 and
look at the value of a to the power of x, we see that
this goes to 1.
So in this case, we'll let a be equal to epsilon and x be
equal to 1/i.
As we can see that as i goes to infinity, the value of x,
which is 1/i, does go to 0.
And therefore, in the limit i going to infinity, the value
of epsilon to the 1 over i-th power goes to 1.
And that shows if we plug this limit into the expression
right here that indeed, the term right here goes to 0 as i
goes to infinity.
And all in all, this implies the probability of Zi minus 0
absolute value greater equal to epsilon in the limit of i
going to infinity converges to 0 for all positive epsilon.
And that completes our proof that indeed, Zi converges to 0
in probability.