so would the following look ok for the first part?
f(x) = {e^-αx^β } {αβ [x^(β-1) } x ≥ 0, α>0, β>0
and for the second part it should look like
f(x) = { (1 + αx)/2 for -1 ≤ x ≤ 1, -1 ≤ α ≤ 1 to start off. apologies for missing the x as it is crucial. do you think i can still use 1+α as a...
cdf to pdf and vise versa
hi
i'm looking for help when going from a cdf function:
F(x) = { 1- e^-αx^β x ≥ 0, α>0, β>0
{ 0 x < 0
to getting the corresponding pdf
also i am looking to do the opposite(pdf to cdf)
for:
f(x) = { (1 + α)/2 for -1 ≤ x ≤ 1, -1 ≤ α ≤ 1
{ 0 otherwise
i'm...
hi
i'm looking for help when going from a cdf function:
F(x) = { 1- e^-αx^β x ≥ 0, α>0, β>0
{ 0 x < 0
to getting the corresponding pdf
also i am looking to do the opposite(pdf to cdf)
for:
f(x) = { (1 + α)/2 for -1 ≤ x ≤ 1, -1 ≤ α ≤ 1...