Convergence in Probability Example Video Lecture

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.