Question

Can someone please show me how to find the marginal distribution of X, from...

Answer 1

\[f(x,y)=3/2x^3y^2, 1/x \le y \le x, 1\le x \le \infty\]

Answer 2

\[f_x(x)=\int\limits_{1}^{\infty}\frac{3}{2x^3y^2}dy\]

Answer 3

@SKMC

Answer 4

thinking...bit slow, bear with me

Answer 5

oky:)

Answer 6

\[\int\limits_{1}^{\infty}x 3/2x^2y^2\]

Answer 7

integrate with respect to y so u treat x as a constant

Answer 8

sorry not great at this notation - the x is supposed to be separate from the 3 over 2x etc

Answer 9

i dont understand can u use brackets please

Answer 10

\[\int\limits_{1}^{\infty} (y) (3/2x^3y^2)\]
\[\int\limits_{1}^{\infty} 3/2x^2 dy\]

Answer 11

\[\int\limits_{1}^{\infty} y \frac{3}{2x^3y^2}dy\]??

Answer 12

yes - I cant make the notation do that!

Answer 13

@skmc what's the lower limit of y
??<y<x

Answer 14

That's what i got but it doesn't make sense to me. The function is correct. It was the limits that confused me. i.e. \[(1/x) \le y \le x\] and\[1\le x < \infty\]

Answer 15

i made a mistake :S sorry

Answer 16

let me do it again

Answer 17

no problem :)

Answer 18

\[f_x(x)=\frac{3}{2x^3}\int\limits_{\frac{1}{x}}^{x}\frac{dy}{y}\]\[\large{=\frac{3}{2x^3}[\ln(y)]_{\frac{1}{x}}^{x}}\]\[=\frac{3}{2x^3}[\ln(x)-\ln(\frac{1}{x})]\]

Answer 19

\[=\frac{3}{2x^3}[\ln(x)-[\ln(1)-\ln(x)]]=\frac{3}{2x^3}(2\ln(x))=\frac{3\ln(x)}{x^3}\]

Answer 20

I'm probably missing something very simple here, but the part I don't understand is how the y^2 just disappears

Answer 21

yu said its \[y \frac{3}{2x^3y^2}\]so one y from the bottom cancels with the y on the top

Answer 22

ah of course. Thank you very much for your help - I really appreciate it.

Answer 23

im sorry for the mistake at the beginning,,
and ur welcome:)

Answer 24

No problem. I find this maths really difficult so i do really appreciate your time and effort :)

Answer 25

id love to help if u need anymore help, and i promise i wont make mistakes hehe .. Goodluck

Answer 26

I actually have a few more questions to do. So i will go through them and you might see me back tomorrow night. Im sure I will need more help so fingers crossed you will be around :) Thanks again

Answer 27

haha i'd love to help ,, and ill make sure i come on tomorro :)

Answer 28

How do i find the marginal distribution of Y? I think I have to integrate with respect to x, but i've drawn a blank!

Answer 29

\[f_y(y)=\frac{3}{2y}\int\limits_{1}^{\infty}\frac{dx}{x^3}=\frac{3}{2y}(\frac{-1}{2x^2})_{1}^{\infty}\]\[=\frac{3}{2y}(0-\frac{-1}{2})=\frac{3}{4y}\]

Answer 30

Hi lalaly
Thanks so much - I will attach my workings - i mucked up the marginal distribution of y...

Answer 31

u gave me the wrong fuction at the beginning,,, U added an extra y in the numerator,, thats y here its 3/4y^2

and that means the marginal of x that i found was wrong :(

do u want me to do it again for u with the right function?

Answer 32

Sorry - that would be great thanks, the function that you have written in your first post back to me is correct

Answer 33

yeah but then u said thers another y .. lol its ok
the marginal of y in ur solutions is right,,, but marginal of x isnt

Answer 34

sorry about that

Answer 35

\[\large{f_x(x)=\int\limits_{\frac{1}{x}}^{x}\frac{3}{2x^3y^2}dy}\]\[\large{=\frac{3}{2x^3}\int\limits_{\frac{1}{x}}^{x}\frac{1}{y^2}dy}\]\[\large{=\frac{3}{2x^2}(-\frac{1}{y})_{\frac{1}{x}}^{x}}\]\[\large{=\frac{3}{2x^3}(-\frac{1}{x}-(-\frac{1}{\frac{1}{x}})}\]\[=\frac{3}{2x^3}(-\frac{1}{x}+x)\]just simplify that

Answer 36

3/2x^3 in the third line,, typo

Answer 37

simplified to 3(x^2-1)/2x^3 ?

Answer 38

no\[\frac{3x^2-3}{2x^4}\]

Answer 39

I'm a bit confused - why is it 3x^2-3 instead of 3x^2-1?

Answer 40

3(x^2-1) is the same

Answer 41

of course it is (very embarrassing)

Answer 42

lol its ok, i can tell ur Tired of studying lol

Answer 43

you have been fabulous (and very patient!). Im doing my masters in statistics and it has been about 15 years since i have tackled this kind of maths so it doesn't come very easy to me!

Answer 44

yeah i understand,,, Im always here to help:D:D

Answer 45

thanks so much. I really appreciate your time and effort. Talk to you soon (I have 2 questions to go lol!)

Answer 46

lol ur welcome, and im looking forward to ur questions:P

Answer 47

Some more workings....

Answer 48

im not sure about the first one ,,, but this how i wouldve done it

\[\int\limits_{2}^{1}0dx+\int\limits_{1}^{\infty} \frac{3x^2-3}{2x^4}dx\]

Answer 49

i dont understand why u put p(x<=2) --> integral from 2 to infinity ... x is less than 2 so -infty to 2

Answer 50

Thanks - i see where i went wrong - I always make this error. I think i have it straight now :)