Question

Give a function how do I find B so that f(x) is a continuous RV pdf?

Answer 1

\[f(x) = \frac{ 1 }{ \sqrt{\pi} } e^{-x^{2} +x -B}\]

Answer 2

I know we need to manipulate the gaussian distribution

Answer 3

so I did \[\int\limits_{-\infty}^{\infty} f(x) = 1\]

Answer 4

And rewrote f(x) as \[f(x) = \frac{ 1 }{ \sqrt{\pi} } \exp(-[(x-0.5)^{2} + B - 0.25)]\]

Answer 5

yep so you need B - .25 = 0

Answer 6

but actually, you forgot to distribute... so it should be
B + .25 = 0

Answer 7

How did you get B - 0.25=0?

Answer 8

another way to look at it:

\[-x^2 +x+B=-x^2 +x -.25 = -(x-.5)^2\Rightarrow B=-.25\]

Answer 9

I don't get it... I thought we had to rewrite it so that it's in the form of a gaussian distribution because then the integral to that distribution would be 1.

Answer 10

Then whatever is left over we can solve for B

Answer 11

\[\frac{ 1 }{ \sqrt{2 \pi \sigma^{2}} } \exp (- \frac {(x-\mu)^{2} } {2\sigma^{2}} )\]

Answer 12

Oh do you want the part a-0.25 to be zero because then we'd be left with a gaussian distribution?

Answer 13

there you go

Answer 14

don't you need the value 2's in the eqn as well?

Answer 15

if you use \[f(x) = \frac{ 1 }{ \sqrt{\pi} }e^{-x^2}\] as the density function, then \[f(x) = \frac{ 1 }{ \sqrt{\pi} }e^{-(x-h)^2}\] will have the same area under the curve as it is just a horizontal shift.

Answer 16

Oh I see, my question is just that there is a sqrt(2) in the original gaussian distribution, would I need to multiply f(x) by sqrt(2)/sqrt(2) to get it in the same form

Answer 17

there is also a 2 in the exponential of the regular gaussian distribution..

Answer 18

http://en.wikipedia.org/wiki/Normal_distribution

have a look at the alternative deifinitions (parameterizations).
of course they still have an area of 1 under the curve (to be a density function and probability distribution) but you get differently shaped cuvres.

Answer 19

Okay I'll take a look thanks for your help

Answer 20

you're welcome