[Script Info]
Title:
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:02.00,Default,,0000,0000,0000,,The correct answer here is no.
Dialogue: 0,0:00:02.00,0:00:06.00,Default,,0000,0000,0000,,Suppose we let "x" go to infinity,
Dialogue: 0,0:00:06.00,0:00:11.00,Default,,0000,0000,0000,,then x-Mu2 for any fixed Mu would go to infinity.
Dialogue: 0,0:00:11.00,0:00:14.75,Default,,0000,0000,0000,,So if x with a -infinity would go to zero--
Dialogue: 0,0:00:14.75,0:00:16.00,Default,,0000,0000,0000,,because it is a constant--
Dialogue: 0,0:00:16.00,0:00:21.00,Default,,0000,0000,0000,,Therefore, we know that in the limit of x goes to infinity--
Dialogue: 0,0:00:21.00,0:00:24.75,Default,,0000,0000,0000,,this expression must be zero.
Dialogue: 0,0:00:24.75,0:00:28.00,Default,,0000,0000,0000,,However, in this graph--it stays at alpha, and doesn't go to zero--
Dialogue: 0,0:00:28.00,0:00:32.00,Default,,0000,0000,0000,,So, therefore, there can't be a valid Mu at sigma square.
Dialogue: 0,0:00:32.00,0:00:35.00,Default,,0000,0000,0000,,If a deep improbability, you know that the area in the Gaussian
Dialogue: 0,0:00:35.00,0:00:37.00,Default,,0000,0000,0000,,has to integrate into one,
Dialogue: 0,0:00:37.00,0:00:41.00,Default,,0000,0000,0000,,and this area diverges, it is actually infinite in size,
Dialogue: 0,0:00:41.00,9:59:59.99,Default,,0000,0000,0000,,so it's not even a valid execution.