Moments of Y from X

thorxes1 · #1 09-22-2008, 01:32 AM

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 $M_Y(t)$

jraven · #2 09-22-2008, 01:44 AM

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 $M_X(t)$ .]

thorxes1 · #3 09-22-2008, 02:13 AM

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

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