Let $X$ be a continuously distributed random variable whose PDF is given by: \begin{align*} f_X(x)=K\times 1[x\in [-1,1]], \end{align*} where $K\in \mathbb{R}$ is a constant that needs to be determined. Find the value of $K$, and charaterize the support of $X$.

So I know that $K=\frac{1}{2}$, I am struggling to express the steps. I want to perform the operation in such a way that I keep the indicator function inside the expression. This is because I also want to charaterize the support.

Here is my attempt \begin{align*} \lim _{x\rightarrow \infty}F_X(x)&=\lim _{x\rightarrow \infty}\int^x _{-\infty} K\times 1[t\in [-1,1]]dt\\ &=\lim _{x\rightarrow \infty}Kt\times 1[t\in [-1,1]]\Big|^x_{-\infty}\\ &=??\\ &=K+K\\ &=2K\\ &=1 \end{align*} I am kind of stuck at step 3, where I am unable to evulate the limit of an indicator function. Finally, it would be extremely helpful if you guys could provide some useful properties of indicator function when evulating its integral and limits. Thank you!

  • $\begingroup$ I think you might be overcomplicating it, $\int_{\infty} ^{\infty} \mathcal{1}_{a,b} \mathrm{d}x = b-a$, do not even need to take limits or so $\endgroup$ – An aedonist Sep 14 '17 at 16:59
  • $\begingroup$ @ An aedonist Hi I would really appreicate a more detail answer. I will upvote it regardless of whether its accepted or not. $\endgroup$ – chuck Sep 14 '17 at 17:01

I think you might be overcomplicating it,

$$\int_{-\infty} ^{\infty} \mathcal{1}_{a,b} \mathrm{d}x = \int_a^b \mathrm{d}x = b-a$$

because the integrand is $0$ outside the interval $(a,b)$ related to the indicator function.

So you do not even need to take limits or so, you can simply immediately restrict the domain of integration.

In the example you give, integrating the PDF over the real line simply gives $K \big(1-(-1) \big) = 2K$, and done.

  • $\begingroup$ I want to keep the $F(X)$ in a form that contains the indicator function. This is because later if I want to characterize the support of (X), I will need the indicator function to be there. $\endgroup$ – chuck Sep 14 '17 at 17:04
  • $\begingroup$ I see, i apologise I thought your point was how to calculate $K$. $\endgroup$ – An aedonist Sep 14 '17 at 17:06
  • $\begingroup$ Let me think about it: the fact is that if you intergrate an indicator function, you get a piecewise linear function: $0$ for $x<a$, then linearly increasing, then $1$ for $x>b$. $\endgroup$ – An aedonist Sep 14 '17 at 17:13
  • $\begingroup$ Hi, I think I figure this out now. First evaluate the integral using your method. Then convert it back to function with an indicator function after integrating it. $\endgroup$ – chuck Sep 14 '17 at 17:57

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