What Is the PDF of X^2 for a Uniformly Distributed Variable X?

[Tamis]

Oke this is a simple question but it has me a bit stumped.

Given a random variable X with a uniform probability distribution between [0,2].

What is the probability distribution function (pdf) of X^2 ?

 

[Stephen Tashi]

Caclulate the cumulative distribution of X^2 by integration. Then find it's pdf by differentiating its cumulative.

 

[Mandelbroth]

Oke this is a simple question but it has me a bit stumped.

Given a random variable X with a uniform probability distribution between [0,2].

What is the probability distribution function (pdf) of X^2 ?

Well, let's think through it logically. According to what you have said, the probability distribution function for $X$ is $P(X=k)=\left\{\begin{matrix} 1/2 & k\in [0,2] \\ 0 & k\not\in [0,2] \end{matrix}\right.$

If $X$ is between 0 and 2, then $X^2$ is between 0 and 4. Since we square the stochastic variable, we square its pdf to attain the probability distribution.

Thus, our new pdf is given by $P(X^2=k)=\left\{\begin{matrix} 1/4 & k\in [0,4] \\ 0 & k\not\in [0,4] \end{matrix}\right.$.

 

[Stephen Tashi]

we square its pdf to attain the probability distribution.

That isn't correct. Do the integration instead.

P(X^2 \le x) = P( 0 \le X \le \sqrt{x}) since, in this particular problem, X has zero probability of being in [-\sqrt{x},0].
For 0 \le x \le 4 , the cumulative is given by:
P(X^2 \le x) = \int_0^{\sqrt{x}} \frac{1}{2} dx

 

[Tamis]

Thnx Stephen! that was exactly what i was looking for!

 

[Mandelbroth]

That isn't correct. Do the integration instead.

P(X^2 \le x) = P( 0 \le X \le \sqrt{x}) since, in this particular problem, X has zero probability of being in [-\sqrt{x},0].
For 0 \le x \le 4 , the cumulative is given by:
P(X^2 \le x) = \int_0^{\sqrt{x}} \frac{1}{2} dx

That makes sense...but why doesn't my answer work? I'm off by a factor of $\frac{1}{\sqrt{x}}$.

Edit: Derp. The square root of a positive real number greater than 1 is a smaller positive real number. Thus, there is a skew. Sorry. Ignore me while I sit in the corner of shame.

 

[Stephen Tashi]

This problem is an example of the fact that the pdf f(x) of a continuous random variable X doesn't really have the interpretation "f(x) is the probability that X = x".

If it did have that interpretation then we could claim "the probability that X^2 = x) = the probability that X = sqrt(x) or x = - sqrt(x)), which in this problem is the probability that X = sqrt(x), which is the constant 1/2 for x in [0,4]".

Even though thinking about a pdf in such a manner gives the wrong answer in this problem, there are many other situations in probability theory where thinking about the pdf in that wrong way does suggest the correct formula.

 

[NegativeDept]

Another way to solve this type of problem is the change-of-variables rule for PDFs.

The method Stephen Tashi described is also useful. IMO it is a very good idea to do some simple practice problems (like this one) with both methods.