Question

Find variance of uniform distribution on [0,1] using moment generating function

Answer 1

\[M(t) = E(e^{tx}) = \int\limits_{0}^{1}e^{tx}dx = \frac{e^{t}-1}{t}\]
I checked this and it's correct.
Now then:
\[M'(t) = \frac{te^t-e^t+1}{t^2}\]
so
\[E(X) = M'(0) = \lim_{t \rightarrow 0}\frac{te^t-e^t+1}{t^2} = \frac{1}{2}\]

Variance formula is \(E(X^2) - E(X)^2\)

And \[E(X^2) = M''(0)\]
and we calculate \[M''(t) = \frac{t^2e^t-2te^t+2e^t-2}{t^3}\]
and
\[M''(0) = \lim_{t \rightarrow 0}\frac{t^2e^t-2te^t+2e^t-2}{t^3} = 0\]

But then
\[E(X^2) - E(X)^2 = 0 - \frac{1}{4} = -\frac{1}{4}\]

which can't be right.
I can show my work for every step but it takes too long to type so if there's one part where you think it's wrong please tell me.

Answer 2

Hopefully I can check your computational work. Let me take a look

Answer 3

BRO I AM AN IDIOT I calculated M''(0) wrong

Answer 4

M''(0) is 1/3 not 0

Answer 5

basically I was trying to use l'hospital's rule but instead of doing the derivative of top and bottom, I did the derivative of the entire thing instead

Answer 6

Unless i'm wrong why did you take the limit of M'(t) approaching 0 to find M'(0) if you had M'(t)? Other than that the lhopitals seems good

Answer 7

Oh lmao I was going to ask

Answer 8

cuz it's undefined at 0

Answer 9

Oh yeah got it. didn't see there

Answer 10

Okay..I got the same for the second derivative. let me see the M''(0)

Answer 11

I got 1/3 ur right

Answer 12

Is it supposed to be 1/3? I don't know the stat part, I don't think AP stat is enough for this

Answer 13

Nah AP stats is not enough. You also need calculus

Answer 14

As long as you learn how to use this symbol: \[\int\limits_{a}^{b}f(x)dx\] then that's enough

Answer 15

definitely bruv