Proof: Integral of PDF of Normal Distribution is Equal to 1 (in English)

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Computation Empire

Computation Empire

Күн бұрын

This video shows how to prove that the Integral of PDF of Normal Distribution from negative infinity to positive infinity is Equal to 1 explained in English.
References:
Evaluation of Integral of e^(-x^2) or the Gaussian Integral: • Integral of e^(-x^2). ...
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Пікірлер: 17
@sagessendulani
@sagessendulani 3 жыл бұрын
You save my life with this video. Thank you very much
@halimamotee9247
@halimamotee9247 3 жыл бұрын
Extremely helpful, thanks
@churunlmaknun2501
@churunlmaknun2501 3 жыл бұрын
Thank you, it helps a lot
@andorLOLChannel
@andorLOLChannel Жыл бұрын
Very explanation and understandable helpful!
@muhamadrifai3216
@muhamadrifai3216 3 жыл бұрын
Thanks for the proof, Sir!
@subharthipradhan5987
@subharthipradhan5987 2 жыл бұрын
Sir, can you provide table value through pdf manually????
@TGSG-rb5hz
@TGSG-rb5hz 3 жыл бұрын
God bless you
@융융-o3k
@융융-o3k 3 жыл бұрын
Thanks! I perfectly understand👍
@saqihere
@saqihere 3 ай бұрын
Thank
@tomgreen5736
@tomgreen5736 4 жыл бұрын
when you take the square root of z^2 shouldn't it be abs(z)? Is there some way you know z is always positive?
@computationempire8603
@computationempire8603 4 жыл бұрын
If you are referring to y^2 = z^2/2. I'm showing that to show my objective of simplification. I can either use y = z/sqrt(2) or y = -z/sqrt(2) as solutions to that equation. So y can either be positive or negative. For the first one, it results in the integral limits of from -infty to +infinity since as z -> infinty, y -> infinty (and also with negative infty). For the second one, it results in the integral limits of from +infinity to -infinity but dy will be negative. It should result in the same answer. So in conclusion, you can choose either y=z/sqrt(2) or y =-z/sqrt(2). Just the first one is simpler.
@ymstatistics2023
@ymstatistics2023 Жыл бұрын
👏👏
@owusualfredkwabena8929
@owusualfredkwabena8929 2 жыл бұрын
Thanks Sir
@zhangshijia1752
@zhangshijia1752 Жыл бұрын
Hello, I am a little confused about the basics, why -(sigma * Z)^2 = -sigma^2 * Z^2
@MikaelSepaul
@MikaelSepaul 6 ай бұрын
Look: -(a * b)^2 = - [ (a)^2 * (b)^2] The negative sign is applied to both factors. So, like the exponential to the second power, since both are in the same parenthesis, each of the items in the parenthesis is squared first and then the negative sign is changed.
@dancus25
@dancus25 Жыл бұрын
a jövőkor figyelmébe: ennek a videónak a kibővített verzióját elfogadja Soós néni
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