Verified answer

You need to check for two things:

probability of any individual event lies between 0 and 1 (no negative probabilities and no probabilities that exceed 100%)

Integral of pdf = 1.000

f(x) = 1/3(1-x^2) on [0, 1]. Observe that 1/3(1-x^2) > 0 for all x in the interval. Observe the maximum value of f(x) = 1/3 (i.e., less than 1.00), so it is a good candidate for a pdf.

Finally, Integral f(x) = 1/3x - 1/9x^3 from 0 to 1. F(1) - F(0) = 1/3 - 1/9 - 0 + 0 = 2/9. FAIL. It would have been if the function were g(x) = 9/2 f(x).

Likewise, f(x) = e^x is positive for all x. e^(0) = 1, but e^(ln 2) = 2 > 1, so it cannot be.

Integral of e^x = e^x from 0 to ln 2 = 2 - 1 = 1. (If not for the fact there are events with probabilities exceeding 1.00, this might have been a pdf.

If any function f(x) is a PDF on some interval [a,b] then

Integral[a,b](f(x) dx) = 1

Let f(x) = 1/3 (1-x^2) on [0,1]

Integral[0,1](f(x) dx) = 1/3*(x -1/3 x^3)[0,1] = 1/3*{1 - 1/3} = 2/9 so this is not a PDF

Let f(x) = e^x on [0, ln2]

Integral[0,ln(2)](f(x) dx) =e^x[0,ln(2)] = {1 -e^(ln(2))} = 1 - 2 = -1 again not a PDF

(1/3) ∫ (1-x^2) dx

= (1/3) ( x-x^3/3)

= x/3 - x^3/9

Let F(x) = x/3 - x^3/9

F(1) = 1/3-1/9 = 2/9

F(0) = 0

F(1)-F(0) = 2/9

Since f(x) does not integrate to 1 on (0,1), f(x) is not a probability density function.

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∫ e^x dx = e^x

Let F(x) = e^x

F(ln 2) = e^ln 2 = 2

F(0) = e^0 = 1

F(ln 2) - F(0) = 2-1 = 1

Since the function integrates to 1 on [0, ln2], f(x) is a probability function.

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