0:00:00.000,0:00:02.000
The correct answer here is no.
0:00:02.000,0:00:06.000
Suppose we let "x" go to infinity,
0:00:06.000,0:00:11.000
then x-Mu2 for any fixed Mu would go to infinity.
0:00:11.000,0:00:14.750
So if x with a -infinity would go to zero--
0:00:14.750,0:00:16.000
because it is a constant--
0:00:16.000,0:00:21.000
Therefore, we know that in the limit of x goes to infinity--
0:00:21.000,0:00:24.750
this expression must be zero.
0:00:24.750,0:00:28.000
However, in this graph--it stays at alpha, and doesn't go to zero--
0:00:28.000,0:00:32.000
So, therefore, there can't be a valid Mu at sigma square.
0:00:32.000,0:00:35.000
If a deep improbability, you know that the area in the Gaussian
0:00:35.000,0:00:37.000
has to integrate into one,
0:00:37.000,0:00:41.000
and this area diverges, it is actually infinite in size,
0:00:41.000,9:59:59.000
so it's not even a valid execution.