density functions

qwertykeyboards · 13-01-2009 10:04PM #1

Hi, im havin trouble understanding how the joint density is achieved in this question. i have the total solution to the problem but cant understand where step 1 gets pulled out of, is there a formula i should know about or what? any help really appreciated.

heres the question.

"Assume that V ~ X^2_r and W ~ X^2_s and that these random variables are independent. Determine the density function of

F = (V/r)/(W/s).

Solution begins like this...

step 1: Joint density of V and W is

{[v^((r/2) - 1) . e^-v/2] . [w^((s/2) - 1) . e^-w/2]} / {[2^(r/2) . R (r/2)] . 2^s/2 . R (s/2)]}

= kv^((r/2)-1) w^((s/2) -1) . e^(-1/2(v+w)) , k = 1/ [2^(r/2) R (r/2) 2^s/2 . R (s/2)]

hope u can understand that. thanks a million!

Grudaire · 13-01-2009 10:36PM

The density of V and W is the Chi-squared density, ie

[v^((r/2) - 1) . e^-v/2] / [2^(r/2) . G (r/2)] - G = gamma

so the joint density is f(V,W) = h(V) . g(W) where:

h(V) = density of V
g(W) = denity of W

which gives you f(V,W) = your function

qwertykeyboards · 14-01-2009 11:56AM

i looked up the chi-squared prob dens function and understand how this bit is achieved,

qwertykeyboards wrote: »

{[v^((r/2) - 1) . e^-v/2] . [w^((s/2) - 1) . e^-w/2]} / {[2^(r/2) . R (r/2)] . 2^s/2 . R (s/2)]}

but im still unsure where this bit comes from.

qwertykeyboards wrote: »

= kv^((r/2)-1) w^((s/2) -1) . e^(-1/2(v+w))

this question alone could nearly pass me on this exam so just want to be sure on everythin! thanks.

Grudaire · 14-01-2009 01:50PM

{[v^((r/2) - 1) . e^-v/2] . [w^((s/2) - 1) . e^-w/2]} / {[2^(r/2) . R (r/2)] . 2^s/2 . R (s/2)]}

= {v^((r/2) - 1) * w^((s/2) - 1) * e ^-v/2 * e^-w/2} * 1/{[2^(r/2) . R (r/2)] . 2^s/2 . R (s/2)]}

let k = 1/ [2^(r/2) R (r/2) 2^s/2 . R (s/2)]

= k * v^((r/2)-1) w^((s/2) -1) . e^(-1/2(v+w))

it's just simplification where you take out k as a constant so you don't have to write the whole long thing

qwertykeyboards · 14-01-2009 03:15PM

thanks for the help as usual mate! maths stats test gonna be a load of balls!

qwertykeyboards · 14-01-2009 07:29PM

sorry man just 2 more problems, firstly, in the last bit of the solution (its the last page of keoghs notes) where we're solving h (f) he changes the integral from 0 up to infinity of [x^((r+s)/2)] . (e^(-x)) (dx) into G ((r+s)/2) G = Gamma

i cant see where he's getting this from. nothing else changes on the 2 lines except for this.

also, in the other density function problem, the one on the previous page of his notes (8.1), in step 1 he gets pi into his joint density function. what is that all about? is it somethin to do with the Z ~ N (0, 1) at the top? and what does that mean if its not?

Grudaire · 14-01-2009 10:11PM

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

It's the same thing

and yea, the density of Z is e^(-(z^2)/2)/sqrt(2*pi) times something if Z ~ N(0,1)

qwertykeyboards · 15-01-2009 11:52AM

cliste, this is the only question 5 that differs from the other 2 kinds in the last 9 years, do you know how this one is done differently? he'd be actin the bollo* to put it up hasnt been up in 7 years or somethin. have the other 2 learned now thanks for ur help with them.

Grudaire · 15-01-2009 08:35PM

qwertykeyboards wrote: »

cliste, this is the only question 5 that differs from the other 2 kinds in the last 9 years, do you know how this one is done differently? he'd be actin the bollo* to put it up hasnt been up in 7 years or somethin. have the other 2 learned now thanks for ur help with them.

Same strategy:

Joint density X,Y

Find jacobian and Change variables,

integrate with respect to the second variable

and you've your density

density functions

Comments