00:01
So in this question we have x which takes the values minus 2, minus 1, 0, 1, and 2, each with the probability of a fifth, and y is x squared.
00:16
So let's do the joint distribution of x and y.
00:19
X can take values minus 2, minus 1, 0, 1, or 2, and y can take values 0, 1, or 4.
00:32
So let's tabulate this.
00:42
Now x can take value minus 2 with probability of fifth, but if x is minus 2 then x squared is 4.
00:51
X is minus 1 with probability of fifth, but if x is minus 1 then x squared is 1.
00:57
If x is 0 then x squared is also 0.
01:01
If x is 1 then x squared is 1, and if x is 2 then x squared is 4.
01:06
So that's our probability distribution.
01:13
Now let's actually do marginal distributions as well.
01:18
P x we just sum along the x columns, and let's do a y marginal as well.
01:37
Now what we can do to test independence is that they're independent if for all x and y p x y equals p x times p y.
01:54
So for example p x equals 0, y equals 1 is equal to 0, because we're in this box here, but p x equals 0 times p y equals 1 is equal to 1 fifth times 2 fifths.
02:18
So that's 2 over 25...