Expectation of Random Variables examples

Question

I am reading Intro to Math Stats by Hogg, McKean & Craig and I've come across an example that I don't understand, and I am not sure if it's a mistake in the text or not. I disagree with the bold part:

Let $X$ be a continuous random variable with the pdf $f(x) = 2x$ which has support on the interval $(0, 1)$. Suppose $Y = \frac{1}{(1+X)}$. Then,

$E(Y) = \int_0^{1} \frac{2x}{1+x}dx$.

The support of a continuous random variable $X$ was previously defined in the text as all $x$ such that the $f_X(x) > 0$. Then in the above example, why isn't the support $(0, \infty)$? And the integral then would be from $0 $ to $\infty$ as well, no?

Some explanation appreciated :)

Answer 1

The support of a random variable depends on the context of the problem. For example, you could have a random variable $X$ with the property that it has support on $(0, \infty)$, but it would need to be appropriately normalized in order to be a valid PDF. That is, a random variable with support $\Omega$ must have the property that $\int_{\Omega} f_X (x) \ dx = 1$.

Consider your density, $f_X (x) = 2x$, and suppose for a contradiction that it has support on the entire positive real line. Then: \begin{align*} \int_{\Omega} f_X (x) \ dx = \int_0^{\infty} 2x \ dx = \frac{2x^2}{2} \Bigg|_0^{\infty} = x^2 \Bigg|_0^{\infty} = \infty. \end{align*} This is certainly not $1$, a contradiction. Suppose instead that it has support on $(0,1)$: \begin{align*} \int_{\Omega} f_X (x) \ dx = \int_0^1 2x \ dx = \frac{2x^2}{2} \Bigg|_0^1 = x^2 \Bigg|_0^1 = 1, \end{align*} which is exactly right and the PDF is valid.

Answer 2

Pdfs should satisfy $\int f(x)\,dx=1$. You can say that $f$ has support in $(0,\infty)$ but in this case, to make sure the above integral valid, we need to determine where $f(x)$ becomes $0$. That will make the problem unclear. When you calculate $E(Y)=\int \frac{f(x)}{1+x}\,dx$, your integral domain can be $(0,\infty)$, but you should note that $f=0$ outside $(0,1)$.