00:01
Okay, given two special random variables, we are required to compute the covariance between them, and then conclude that although the covariance between them is equal to zero, they are actually dependent with each other.
00:23
They are not independent.
00:24
Okay? so, let's say y is a random variable with density function, let's say pdf symmetric, symmetric, about zero.
00:43
Then from this definition, it's easy for us to know.
00:47
The expectation of y is equal to zero.
00:50
Okay, now x is some random variable dependent by y.
00:58
Okay, let's say s is equal to sy, and s is to y.
01:06
And the probability of s equal to one is equal to the probability of s being negative one, and those two probabilities are both equal to one half.
01:24
Okay, now let's consider the covariance.
01:29
By the definition, the covariance of x and y is equal to the expectation of xy minus expectation of x times the expectation of y.
01:44
Notice, we know the expectation of y is zero, so we know that this product is equal to zero.
01:52
So to show this covariance, to show the fact that this covariance is equal to zero, we only need to show this expectation is equal to zero.
02:05
Or equivalently, we want to show the expectation of x, s times y squared is equal to zero.
02:21
Okay, now as s and y are different random variables to compute the whole expectation, we can use the method called the conditional expectation.
02:33
I mean, the whole expectation can be written as the expectation with respect to y is the conditional expectation, y squared, conditioning on s, times the condition, the expectation with respect to s.
02:56
As this random variable will just take two values, the first one is one, while the second one is minus one.
03:06
So we can consider this expectation first by the definition.
03:10
It is equal to the expectation.
03:16
Now, let's see if you can write it in this way.
03:23
This equals the probability s equal to one.
03:28
Now, as under the condition that s is equal to one, we know this whole thing is equal to, s is equal to one, so expectation y squared.
03:46
Okay, plus the probability s equal to s being minus one, expectation.
03:53
Now, as under this condition, we know this should be minus y squared.
04:06
So we'll get one half times expectation of y squared, plus one half times expectation of minus y squared.
04:17
And by the linearity of the expectation, we know this dimension is equal to zero.
04:22
That means, okay, the expectation of xy is just equal to this guy, and it's equal to zero.
04:30
And from our discussion, we know the covariance must be equal to zero.
04:41
Okay, now we just do some discussion to show xy are not independent.
04:48
In fact, from the expression of x and y, because x can be expressed as s times y...