How do you find moment generating function?

[semidevil]

I have absolutly no idea how to do this.

so let X be a random variable with pdf fx(xy) =
x for 0<=x<=1
2 - x for 1 <= 1 <= 2
0 otherwise.

I"m looking through my book, and it doesn't give examples that resembles this.

all I see is the moment is e^(tk) * the function...

and tI don't know what to do when it comes to my problem.

is it the integeral from 0 to 1 of e^(tk) * x + the integeral from 1 to 2 of e^tk * 2 - x?

 

[Dr.ThinkDeep]

If you mean int(e^(t*k) *x,x= 0 .. 1) + int(e^(t*k) * (2-x),x=1 ..2) you are right.

 

[tacman]

Finding a moment generating function involves finding the expected value of e^(tx), where x is the random variable and t is a parameter. This can be done by integrating e^(tx) with respect to x over the range of possible values for x. In this case, we have two different ranges for x: 0 to 1 and 1 to 2.

To find the moment generating function for this specific random variable, we can break it into two parts:

1. For x values between 0 and 1, the pdf is simply x. Therefore, the moment generating function for this range is:

M(t) = ∫e^(tx) * x dx from 0 to 1

= (1/t) * e^(tx) * x from 0 to 1

= (1/t) * (e^t - 1)

2. For x values between 1 and 2, the pdf is 2-x. Therefore, the moment generating function for this range is:

M(t) = ∫e^(tx) * (2-x) dx from 1 to 2

= (1/t) * e^(tx) * (2-x) from 1 to 2

= (1/t) * (e^(2t) - e^t)

To find the overall moment generating function for the random variable X, we can add these two parts together:

M(t) = (1/t) * (e^t - 1) + (1/t) * (e^(2t) - e^t)

= (1/t) * (e^(2t) + e^t - 1)

So, the moment generating function for this specific random variable is (1/t) * (e^(2t) + e^t - 1). I hope this helps! If you're still having trouble, it may be helpful to review some examples in your textbook or consult with your instructor for further clarification.