Question

For t∈R, define the following two functions:
f1(t)=12π−−√exp(−max(1,t2)2)


and
f2(t)=12π−−√exp(−min(1,t2)2).


In this problem, we explore whether these functions are valid probability density functions.

Determine whether the function f1 is a valid PDF for a continuous random variable that can take any value on the real line. If not, is there a number c>0, such that cf1 is a valid PDF?

Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.
No, it is not a valid PDF, and there is no constant c making cf1 a valid PDF.
None of the above.
unanswered

Determine whether the function f2 is a valid PDF for a continuous random variable that can take any value on the real line. If not, is there a number c>0, such that cf2 is a valid PDF?
Yes, it is a valid PDF.
No, it is not a valid PDF, but there is a constant c making cf a valid PDF.
No, it is not a valid PDF, and there is no constant c making cf2 a valid PDF.
None of the above.

Answer 1

clearly f2 is not a valid PDF, since for |t|>1, f1(t) = k/√e so the area under the curve outside [-1,1] is unbounded.

similarly, f1 is not a PDF, but since it has a limit, a suitable c can be found to make the ∫[-∞,∞] f2(t) dt = 1

Answer 2

sorry -- typo. I meant
∫[-∞,∞] c*f1(t) dt = 1

Answer 3

where are you getting your limits of integration from?

Answer 4

how are the two cases different? in either f1(t) or f2(t), you have to try each case, ie, 1> t^2, t^2> 1 so both results, for f1(t) and f2(t) would be the same, no?

Answer 5

a) No, it is not a valid PDF, but there is a constant c making cf1 a valid PDF.
b) No, it is not a valid PDF, and there is no constant c making cf2 a valid PDF.