Probability density function of a random number

[Proggy99]

Homework Statement

Let X be a random number from (0,1). Find the probability density function of Y = 1/X.

Homework Equations

The Attempt at a Solution

I keep thinking this is easier than I am making it out to be, but the only places I find anything similar searching is on two exams at two different schools, so it must be more interesting than I thought.

I am attempting to first find the distribution function and then differentiate to get the density function (not sure that is the right way to approach)

for any x between 0 and 1, the probability distribution would be
0 if x < 0
x if 0 \leq x < 1
1 if x \geq 1

for y = 1/x
x = 1/y
so F(y) = 1/y and F'(y) = f(y) = -1/y^{2}

I know the answer is 1/y^{2} and not negative, but I have the suspicion that I am doing it wrong anyway and my answer is just coincidentally close to correct. Please help

 

[Focus]

You can tell F(y)=1/y is wrong by taking y=1/2. Consider \mathbb{P}(Y=k) for some k. Now try to get the expression in terms of X.

Good luck.

 

[statdad]

Two approaches.
First, if Y = 1/X, then you are correct that 1 \le Y &lt; \infty.
Now

<br /> P(Y \le y) = P(1/X \le y) = P(X \ge \frac 1 y) <br />

You forgot to reverse the inequality. Write out a form for P(X \ge 1/y) and differentiate that to get the density.

Second: if you want to do a change of variables, you need to multiply be the absolute value of the derivative.

 

[D H]

Let X be a random number from (0,1).

Just picking a nit here, but that should be U(0,1), not just (0,1).

so F(y) = 1/y

That is not correct. This fails two key tests of whether a function is a CDF. This function is monotonically decreasing for y>1 (CDFs must be monotonically increasing functions) and F(y)→0 as y→∞ (CDFs must have F(y)→1 as y→∞).

You are on the right track, however.

 

[Averagedude]

PF Mentor,
Can you post the complete answer?