I don't have answers to refer to, so could please someone check if the way I solved this problem is correct? I have an exam tomorrow and want to make sure that I've got it right. THank you in advance!
Check whether the following functions are probability densities on the corresponding sample space Ω = I:
(a) $f(x) = 1/x, I = [1, e]$, where e = exp(1)
$f(x)>0$ for all x on the $I= [1, e]$
$\int^e_11/xdx=lne-ln1=1$ so it is a PDF.
(b) $f(x) = exp(x/3)$, $I = [0, 1]$
$f(x)>0$ for all x on $I = [0, 1]$
$\int^1_0e^{x/3}=\int^1_0x/3lne=[x^2/6]^{x=1}_{x=0}=1/6 - 0=1/6$ so NOT a pdf
(c) $f(x) = x/4, I = [−1, 3]$
$x/4>0 -> x \not=0$ or any negative number but the interval [-1,3] contains 0 and negative numbers--> f(x) is NOT a pdf
(a) looks correct.
(b) You want to check that again. The anti-derivative of $e^{x/3}$ is $3e^{x/3}$.
(c) $\frac{x}{4} > 0$ does not imply $x > 4$. But you are right that the function is not positive.
EDIT: Ok, you noticed (c) yourself. Good.