Question

What is Integration of Probability Density function ?

Answer 1

for which distribution?

Answer 2

\[\int \frac{e^{-\ \cfrac{(x-mu)^2}{2 \sigma^2}})}{\sqrt{2 \pi} \sigma}\]

Answer 3

spurious ) lol

Answer 4

The probability for a variable to fall within a particular region is given by the integral of this variable’s density over the region. The integral for the entire region is 1

Answer 5

\[\int \frac{e^{-\ \cfrac{(x-\mu)^2}{2\ \sigma^2}}}{\sqrt{2 \pi} \ \sigma}\]

Answer 6

@ amistre how to prive that the given integration is =1

Answer 7

*prove

Answer 8

integrate it from -inf to inf if you want to prove it :)

Answer 9

id have to recall what is considered the variable of integration tho

Answer 10

its "x" ... duh

Answer 11

Yes from - inf to +inf need to prive it as 1

Answer 12

if we adapt this to a mean of 0 and a sd of 1 we get:

\[\int_{-\infty}^{\infty} \frac{e^{-\ \cfrac{(x-1)^2}{2}}}{\sqrt{2 \pi}}dx\]
\[\frac{1}{\sqrt{2 \pi}}\ \int_{-\infty}^{\infty}\ e^{-\ \cfrac{(x-1)^2}{2}}dx\]

oww my head hurts!!

Answer 13

lol .... id have to have wolfram do it

Answer 14

if i tried it it would turn out like it does in the books; fill tomes

Answer 15

id prove it, if need be; as half; from 0 to inf; or at least as the limit of that

Answer 16

youd still have to determine a decent way to integrate it tho

Answer 17

\[\frac{1}{\sqrt{2 \pi}}\ \int_{0}^{\infty}\ e^{-\ \cfrac{(x-1)^2}{2}}dx\]

at least comes from:

\[-A * B\ e^{-\ \cfrac{(x-1)^2}{2}}\]
where A = the front stuff and B = the stuff that makes it work

Answer 18

if i were to derive this as is; the trick is in the exponent itself
\(-\cfrac{1}{2}(x-1)^2\) derives to; \(-\cfrac{2}{2}(x-1)(1)\)

to give us: \(-(x-1)\ e^{C}\); now B has top be a product or a quotient rule to apply if i see it right to try to cancel this out

Answer 19

if you are looking for a proof...you can write it as a double integral and switch to polar coordinates

Answer 20

that would most likely be easier :)

Answer 21

Yes I got it thnks a lot for your help really appreciate

Answer 22

there is no antiderivative for \[e^{-x^2}\]

in terms of elementary functions...therefore writing as a double integral is the way to go.

Answer 23

is there one for e^(x^2) ?

Answer 24

no