Actuarial Outpost Moments of Y from X
 Register Blogs Wiki FAQ Calendar Search Today's Posts Mark Forums Read
 FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions

 D.W. Simpson & Company International Actuary Jobs   Canada  Asia  Australia  Life  Pension  CasualtyBermuda, United Kingdom, Europe, Asia, Worldwide

#1
09-22-2008, 01:32 AM
 thorxes1 Member Join Date: Sep 2008 Posts: 247
Moments of Y from X

Ok so I understand the basic principle of:
$M_Y(t)= e^{bt}M_X(at)$

and with a Bernoulli trial you can just calculate $M_X(at)$ easily enough since there's only 2 probabilities.

I'm running into an issue when probabilities aren't given at all.

I.E. $M_X(t)= e^{5t+2t^{2}}$ and $Y = \frac{(X-5)}{2}$

So first off, it helps to have Y in the form of: ax+b, so I distributed the 1/2 and got :$Y = \frac{1}{2}X- \frac{5}{2}$

ok so now a = 1/2 and b = -5/2.

So I set it up as: $M_Y(t) = e^{bt} * M_X(at)$
= $e^{bt} * E(e^{xat})$
= $e^{bt} * \sum_{x=1}^\infty e^{atx} * p(x)$

=$e^{\frac{-5}{2}t} * \sum_{x=1}^\infty e^{\frac{1}{2}tx} * p(x)$

But how do I calculate $\sum_{x=1}^\infty e^{\frac{1}{2}tx} * p(x)$ ?
I'm not told that it's a bernoulli trial, or how many possible values for X there are.

And on a side note, do I start at x = 0 or x = 1?

And how do you know when to do either or?

The answer is apparently $e^{\frac{t^{2}}{2}$ for $M_Y(t)$
#2
09-22-2008, 01:44 AM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,262

Quote:
 Originally Posted by thorxes1 I'm running into an issue when probabilities aren't given at all. I.E. $M_X(t)= e^{5t+2t^{2}}$ and $Y = \frac{(X-5)}{2}$ So first off, it helps to have Y in the form of: ax+b, so I distributed the 1/2 and got :$Y = \frac{1}{2}X- \frac{5}{2}$ ok so now a = 1/2 and b = -5/2. So I set it up as: $M_Y(t) = e^{bt} * M_X(at)$
Which means
$M_Y(t) = e^{-\frac{5}{2} t} \,\cdot\, M_X\left(\frac{t}{2}\right)$

$M_Y(t) = e^{-\frac{5}{2} t} \,\cdot\, e^{5 \left(\frac{t}{2}\right) + 2 \left(\frac{t}{2}\right)^2} = e^{t^2/2}$

[The second factor comes from plugging t/2 in for t in the formula for $M_X(t)$.]
__________________
The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.
#3
09-22-2008, 02:13 AM
 thorxes1 Member Join Date: Sep 2008 Posts: 247

Wow. I'm so slow. In retro spec I can't believe I missed that connection. Thanks for the kick to the head! <3

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off

All times are GMT -4. The time now is 09:36 AM.

 -- Default Style - Fluid Width ---- Default Style - Fixed Width ---- Old Default Style ---- Easy on the eyes ---- Smooth Darkness ---- Chestnut ---- Apple-ish Style ---- If Apples were blue ---- If Apples were green ---- If Apples were purple ---- Halloween 2007 ---- B&W ---- Halloween ---- AO Christmas Theme ---- Turkey Day Theme ---- AO 2007 beta ---- 4th Of July Contact Us - Actuarial Outpost - Archive - Privacy Statement - Top