Question

find the mgf of f(y)=.5e^(-abs(y)).

Answer 1

I know m(t) = E(e^ty)

Answer 2

which is = to the integral of (e^ty)f(y)dy

Answer 3

but I'm having trouble solving

Answer 4

What's the support of the pdf?

Answer 5

I suppose we're to assume it's all real \(y\)?
\[m(t)=E\left[e^{ty}\right]=\frac{1}{2}\int_{-\infty}^\infty e^{ty}e^{-|y|}~dy\]
Let's consider that absolute-valued exponent. When \(y<0\), you have \(|y|=-y\), and when \(y\ge0\), you have \(|y|=y\). So, split up the integral at 0 and you get
\[\frac{1}{2}\left(\int_{-\infty}^0e^{ty}e^{-(-y)}~dy+\int_0^\infty e^{ty}e^{-y}~dy\right)\]

Answer 6

pdf support is -infinity to +infinity

Answer 7

Ah, good, on the right track then.
\[\frac{1}{2}\left(\int_{-\infty}^0e^{ty}e^{y}~dy+\int_0^\infty e^{ty}e^{-y}~dy\right)\]
\[\frac{1}{2}\left(\int_{-\infty}^0e^{y(t+1)}~dy+\int_0^\infty e^{y(t-1)}~dy\right)\]
\[\frac{1}{2}\left(\frac{1}{t+1}\left[e^{y(t+1)}\right]_{-\infty}^0+\frac{1}{t-1}\left[e^{y(t-1)}\right]_0^\infty \right)\]

Answer 8

why is the first part t+1 and the second t-1?

Answer 9

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
Because \(e^{a}e^{b}=e^{a+b}\). In the first integral, the exponents are \(ty\) and \(y\), so \(e^{ty}e^{y}=e^{ty+y}=\color{red}{e^{y(t+1)}}\). The reasoning applies to other integrand's exponent.
\(\color{blue}{\text{End of Quote}}\)

Answer 10

right. I understand the exponents and adding them and their rules. I'm more confused by the difference between the + and -.

Answer 11

The plus and minus come as a result of the absolute value definition, which I used in my second post.

Answer 12

but the |y| can never =-y which is what you wrote? or am i missunderstanding your post?

Answer 13

Regardless of \(y\)'s sign, \(|y|\) is positive, yes, but \(|y|=y\) is not true if \(y\) is negative.

For example, \(|-2|\not=-2\), since \(|-2|=2\). The minus sign in front of the \(y\) "corrects" for this. In this case, \(y=-2\), and so \(-y=2=|-2|=|y|\).

Answer 14

Does that integral converge?

Answer 15

That's what I was afraid of @wio. Checking with Wolfram, and looks like we need to set some restrictions on \(t\).

Answer 16

okay. I think i understand

Answer 17

but i'm still not sure how to continue

Answer 18

Hey, neither do I! I still haven't covered continuous random variables, so I had to refer to my textbook :)

Answer 19

I hate stats! :(

Answer 20

The moment generating function really don't work here.

Answer 21

but that's the question i was asked to do. :/

Answer 22

I have to get some sleep, if you guys can help me out any further please let me know. Thanks for the help!