- #1
nacho-man
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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 (Clapping)
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 (Clapping)
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