Probability density function of a random number

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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 [tex]\leq[/tex] x < 1
1 if x [tex]\geq[/tex] 1

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

I know the answer is 1/[tex]y^{2}[/tex] 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
 

Answers and Replies

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

Good luck.
 
  • #3
statdad
Homework Helper
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Two approaches.
First, if [tex] Y = 1/X[/tex], then you are correct that [tex] 1 \le Y < \infty [/tex].
Now

[tex]
P(Y \le y) = P(1/X \le y) = P(X \ge \frac 1 y)
[/tex]

You forgot to reverse the inequality. Write out a form for [tex] P(X \ge 1/y) [/tex] 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.
 
  • #4
D H
Staff Emeritus
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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.
 
  • #5
PF Mentor,
Can you post the complete answer?
 

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