SOLUTION: Let X be a continuous random variable. Can the function f(x) = 2^(-x) for x > 0 and 0 otherwise be its probability density function? If so, state why. If not, explain why not.

Question 1136713: Let X be a continuous random variable. Can the function f(x) = 2^(-x) for x > 0 and 0 otherwise be its probability density function? If so, state why. If not, explain why not.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

No.

The integral from 0 to infinity of f(x) = 2^(-x) is 1/ln(2) which is about 1.4427.

For f(x) to be a probability density function, the integral has to be equal to 1.

Note f(x) = 2^(-x) could be a probability density function if it were defined only for positive integer values of x, instead of having a continuous domain of x>0:

2^(-1)+2^(-2)+2^(-3)+... = 1/2 + 1/4 + 1/8 + ... = 1.
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