Help Understanding Nasty Integrals in Math Stats Class

[Snarf]

I'm in a mathematical statistics class and it is spanking me. Please help.

I have two questions that will really help me understand things if I get a nice explanation.

1. A random variable Y has the following probability function p(y) = y$$^{2}$$/15 for y = 1, 2, 3. Findthe moment generating function for Y.

What this problem requires is the integration of m(t) = E[e^ty] = $$\int$$ e$$^{ty}$$y$$^{2}$$/15dy integrated from 1 to 3.

I used integration by parts but succeeded in getting something very large and ugly.

The second question is along the same lines:

2. Let Y be a random variable with $$\mu$$$$^{'}_{k}$$=[1 + 2$$^{k+1}$$ + 3$$^{k+1}$$]/6

I need to inegrate m(t) = E[e$$^{ty}$$] = $$\int$$$$e^{ty}$$[1 + 2$$^{k+1}$$ + 3$$^{k+1}$$]/6 dy from 0 to infinity finding the first four terms and indicating the sum continues.

Anyone?

 

[HallsofIvy]

I'm in a mathematical statistics class and it is spanking me. Please help.

I have two questions that will really help me understand things if I get a nice explanation.

1. A random variable Y has the following probability function p(y) = y$$^{2}$$/15 for y = 1, 2, 3. Findthe moment generating function for Y.

What this problem requires is the integration of m(t) = E[e^ty] = $$\int$$ e$$^{ty}$$y$$^{2}$$/15dy integrated from 1 to 3.

I used integration by parts but succeeded in getting something very large and ugly.

Large and ugly? I don't see why you would! Any time you have a power of x times an integrable function, it should be reasonably easy to reduce. Let u= y² and dv= e^tydy. Then du= 2y dy and v= (1/t)e^ty. You now have
$$(1/t)e^{ty}y^2\right|_{y=1}^{3} -(2/t)\int_{y= 1}^3 ye^{ty}dy$$
Use integration by parts again with u= y, dv= e^tydy.

The second question is along the same lines:

2. Let Y be a random variable with $$\mu$$$$^{'}_{k}$$=[1 + 2$$^{k+1}$$ + 3$$^{k+1}$$]/6

I need to inegrate m(t) = E[e$$^{ty}$$] = $$\int$$$$e^{ty}$$[1 + 2$$^{k+1}$$ + 3$$^{k+1}$$]/6 dy from 0 to infinity finding the first four terms and indicating the sum continues.

Anyone?

Are you sure that's what you have? [1+ 2^k+1+ 3^k+1]/6 is a constant! You only need to integrate e^ty.

 

[Snarf]

I see where I went wrong on the first problem. my dv and my v for the first integration by parts was switched. Thanks for clearing that up.

 

[Snarf]

On the second problem it isn't an integral. Its a sum. Here is how I should have written it.

$$\sum$$e$$^{ty}$$[(1+2$$^{k+1}$$+3$$^{k+1}$$)/6]