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I am trying to learn probability and have gotten to the points of transformations. I am now quite confused. While I understand what CDF and PDF are, I have no idea why changing the value of X to a function on X should drastically alter these, or how I should go about determining them. This is the problem statement I'm trying to figure out:

X is a continuous random variable. Find the CDF and PDF of $Y=X^2$ in the expression of CDF FX of X and PDF fX of X.

How do I proceed here, especially since no actual function is given and no distribution specified?

David_Authors_Books
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1 Answers1

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Here's an extreme example of why composing a function with a random variable alters the distribution. Consider the uniform random variable $X\sim U\{-1,1\}$. That is, $\mathbb{P}(X=-1)=\mathbb{P}(X=1)=1/2$. Now, let $f$ be a function which maps everything to zero. That is, $f(x)=0$. Note that $\mathbb{P}(f(X)=0)=1$, yielding a point mass distribution.

If $X$ is a continuous random variable, $$ \Phi_{X}(x)\equiv\mathbb{P}(X\leq x)=\int_{-\infty}^{\infty}\varphi_{X}(x)dx $$ where $\Phi_{X}$ and $\varphi_{X}$ are the CDF and PDF of $X$. The CDF of $Y\equiv X^{2}$ is given by \begin{multline*} \Phi_{Y}(y)\equiv\mathbb{P}(Y\leq y)=\mathbb{P}(X^{2}\leq y) =\mathbb{P}(\left|X\right|\leq\sqrt{y}) =\mathbb{P}(-\sqrt{y}\leq X\leq\sqrt{y})\\ =\mathbb{P}(X\leq\sqrt{y})-\mathbb{P}(X\leq-\sqrt{y}) =\Phi_{X}(\sqrt{y})-\Phi_{X}(-\sqrt{y}) \end{multline*} To get the PDF from the CDF, just differentiate: $$ \varphi_{Y}(y)=\Phi_{Y}^{\prime}(y)=\frac{1}{2\sqrt{y}}\left(\varphi_{X}(\sqrt{y})+\varphi_{X}(-\sqrt{y})\right). $$ Since $\varphi_X$ was not specified, this is as far as you can simplify.

parsiad
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  • The formula you use for the CDF seems to be the same as the formula used to calculate expectation. Are these the same thing? – David_Authors_Books Jul 31 '19 at 16:50
  • They are not the same thing. The expression for the expectation is $\mathbb{E}[X] = \int_{-\infty}^{\infty} x \varphi_X(x) dx$. You should try to understand the motivation behind this and other expressions. It might help to consider the discrete case separately to do so. – parsiad Jul 31 '19 at 16:51
  • What would happen if x were specified as a standard normal distribution? – David_Authors_Books Jul 31 '19 at 16:52
  • You would plug in $\varphi_X(x) = \exp(-x^2/2) / \sqrt{2\pi}$. – parsiad Jul 31 '19 at 16:56
  • @Andrew_Authors_Books: simplified my answer a bit. – parsiad Jul 31 '19 at 17:02