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probability distribution help please

Random variable X follows a half-Normal distribution with parameter σ^2 if it has the same form of pdf as the Normal distribution with expectation 0 and variance σ^2, but only has a range space of the positive real numbers. Derive the pdf of X and, by using an appropriate substitution or otherwise, evaluate its expectation.

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Original post by zulhaq345
Random variable X follows a half-Normal distribution with parameter σ^2 if it has the same form of pdf as the Normal distribution with expectation 0 and variance σ^2, but only has a range space of the positive real numbers. Derive the pdf of X and, by using an appropriate substitution or otherwise, evaluate its expectation.


Go for it.
Reply 2
How will I go about solving it
Original post by zulhaq345
How will I go about solving it


Do you understand the context they give, first of all ?
Reply 4
what do you mean by that?
Original post by zulhaq345
what do you mean by that?


Do you understand the context:

"Random variable X follows a half-Normal distribution with parameter σ^2 if it has the same form of pdf as the Normal distribution with expectation 0 and variance σ^2, but only has a range space of the positive real numbers."
Can somebody help with this question please
15751986188635496328375323714446.jpg Need help with question 1
I have these notes but have no idea how to apply it to the problem.
1575198939868127460584692135433.jpg15751989731044303406945900415496.jpg
Original post by farhanaktar
Need help with question 1


Didn't you ask this question a few days ago ?? It's important you understand what the context is saying otherwise the explanation below might be confusing.

Anyway, a half-normal distribution is exactly as it sounds like; half of the normal distribution, but adjusted as to make the area under equal to 1 (NOTE that only removing half of the bell curve doesn't make sense as then what you end up with is distribution which does not sum all probabilities to 1, but rather to 0.5. Hence an adjustment is needed).

This half-normal distribution has pdf f(x)f(x) that is zero for x<0x<0 and non-zero for x>0x>0. Your job is to determine what it is for x>0x>0.

You can use the fact that f(x)f(x) is closely related to 1σ2πex22σ2\displaystyle \dfrac{1}{\sigma \sqrt{2\pi}} e^{-\frac{x^2}{2\sigma^2}} which is the pdf of the full normal distribution with mean 0. Hence, it is safe to assume that f(x)f(x) is just some scalar multiple AA of this. Therefore, you seek this AA such that


0f(x).dx=0Aσ2πex22σ2.dx=1\displaystyle \int_0^{\infty} f(x) .dx = \int_0^{\infty} \dfrac{A}{\sigma \sqrt{2\pi}} e^{-\frac{x^2}{2\sigma^2}}.dx = 1

and indeed you can use the fact that this must hold when σ=1\sigma = 1 and for that case your lecture notes give you the result you can use:

ex22.dx=2π\displaystyle \int_{-\infty}^\infty e^{-\frac{x^2}{2}}.dx = \sqrt{2\pi}
Reply 10
These are the notes I have in the pdf. Would you mind letting me a hand in this please?
Original post by zulhaq345
These are the notes I have in the pdf. Would you mind letting me a hand in this please?


Not sure why you ask the same question on two accounts, but I am merging the two threads together now.
Reply 12
so the formulas you have provided, what do I do with them then
Original post by zulhaq345
so the formulas you have provided, what do I do with them then


Do you understand where they come from? Use them to determine AA.
15752064997875626778388217920539.jpg is this fine but I dunno like I still will get A = 1
Original post by farhanaktar
is this fine but I dunno like I still will get A = 1


This is all incorrect.

You know that ex22=2π\displaystyle \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} = \sqrt{2\pi}. Hence, since the integrand is even, this is the same as

20ex22=2π\displaystyle 2\int_{0}^{\infty} e^{-\frac{x^2}{2}} = \sqrt{2\pi} and so 0ex22=2π2\displaystyle \int_{0}^{\infty} e^{-\frac{x^2}{2}} = \dfrac{\sqrt{2\pi}}{2}.

Now coming back to your original problem, we need to find AA such that

0A2πex22=1\displaystyle \int_0^{\infty} \dfrac{A}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} = 1

and you can manipulate LHS so that you have

A2π0ex22=1\displaystyle \dfrac{A}{\sqrt{2\pi}} \int_0^{\infty} e^{-\frac{x^2}{2}} = 1

so what must A be ?
(edited 4 years ago)
Oh okay thank you so much. Is a square root of 2pi15752074206516296913489845102169.jpg
Reply 17
I mean A = 2. then what should I do
Original post by zulhaq345
I mean A = 2. then what should I do


Then you found your pdf for the half-normal distribution.

Go ahead and have a go at calculating the expectation.
15752097166673381821710517459666.jpg

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