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i have a simple enough question

Find the MGF of a continuous random variable with the PDF:

f(x) = 2x, 0<x<1

I understand MGF is calculated as:

$$M(S) = \int_{-\infty}^{+\infty} e^{Sx} f(x)dx$$

which would give me

$$\int_{-\infty}^{+\infty} e^{Sx} 2xdx$$

but how would i compute this integral?

edit: scratch that. I think i am on the right track here, someone check?

if Y = aX + b, then

$$M_{y}(S) = E[e^{2S}] = e^{S}E[E^{2Sx}]$$

$$M_{y}(S) = e^S \int_{0}^{1}e^{2Sx}dx

= ... $$

$$= e^{S}(\frac{e^{2S}}{2s} - \frac{1}{2S}) $$

Is this correct?/ Am I on the correct track?

On another note, let's celebrate me getting the hang of latex! Yay

Find the MGF of a continuous random variable with the PDF:

f(x) = 2x, 0<x<1

I understand MGF is calculated as:

$$M(S) = \int_{-\infty}^{+\infty} e^{Sx} f(x)dx$$

which would give me

$$\int_{-\infty}^{+\infty} e^{Sx} 2xdx$$

but how would i compute this integral?

edit: scratch that. I think i am on the right track here, someone check?

if Y = aX + b, then

$$M_{y}(S) = E[e^{2S}] = e^{S}E[E^{2Sx}]$$

$$M_{y}(S) = e^S \int_{0}^{1}e^{2Sx}dx

= ... $$

$$= e^{S}(\frac{e^{2S}}{2s} - \frac{1}{2S}) $$

Is this correct?/ Am I on the correct track?

On another note, let's celebrate me getting the hang of latex! Yay

Last edited: