4
$\begingroup$

The probability density function of a certain random variable is given by $$ \begin{aligned}g(t)&=4 n(n-1)\\&\int_0^1 x(x+t)1(x+t<1)[(x+t)^2\,1(x+t<1)+1(x+t\ge1)-x^2]^{n-2}\, dx\,1(t>0).\end{aligned}$$ I wonder whether this this integral can be written in a simple way (say, in a case wise manner by splitting in two or more integrals). I put this integral in Wolfram alpha but it is saying Wolfram does not know how to interpret the result. I do not have any idea of the notation used in the expression. Kindly guide what this notation means and how I can write the integral without it.

$\endgroup$
3
  • $\begingroup$ It seems like g(t) is actually, g(t,n) and some constraints on n are required, guessing n>2. Do you think that is implied based on context you have ? To break it case-wise, the power of n-2 bracket would require to be expanded. When the square bracket is expanded, the constraint (x+t<1) could be taken out and incorporated into the integral itself by making the limit of x go from 0 to t instead of 0 to 1. $\endgroup$
    – whisperer
    Commented Sep 7, 2020 at 13:49
  • $\begingroup$ @whisperer Ithink you are closer to my source of confusion.So let me tell u the context .$g(t,n)$ is the pdf of the range of $n$ randomly and uniformly points in a unit circle.The cdf of the radial distance of a random point in a unit circel is given by $$ F(x)=x^2\,1(0<x<1)+1(x\ge1)$$ and formula for the pdf of the range is given by $$ g(t)=n(n-1)\int_{-\infty}^\infty f(x)[F(x+t)-F(x)]^{n-2} f(x+t)\, dx\;1(t>0).$$Of course , $n$ is greater than 2. Your comment has simplified my question.All I actually want is to write the expression in a way which is free from indicator functions $\endgroup$ Commented Sep 7, 2020 at 17:04
  • $\begingroup$ @ I mean the expression in the question is to be freed from indicator functions $\endgroup$ Commented Sep 7, 2020 at 17:13

2 Answers 2

1
$\begingroup$

The initial "$1$" in the notation "$1(x+t\lt 1)$" is a sloppy but suggestive and fairly common shorthand for the indicator function $\mathscr{I}.$ (Usually the "$1$" would be in bold type, "$\mathbf{1},$" to distinguish it from the constant $1.$ As you can see, even here on CV the distinction is so subtle as to be easily overlooked.)

By definition, when $x+t \lt 1,$ $\mathscr{I}(x+t \lt 1) = 1$ and $\mathscr{I}(x+t \ge 1) = 0.$ Otherwise, when $x+t \ge 1,$ $\mathscr{I}(x+t \lt 1) = 0$ and $\mathscr{I}(x+t \ge 1) = 1.$

Here's the only analysis we need: when $t\ge 1,$ $\mathscr{I}(x+t\lt 1) = 0$ for $0\le x\le 1,$ reducing the integral to $0.$ Otherwise, $0 \le t \lt 1,$ implying $\mathscr{I}(t\gt 0)=1.$

Break the integral at the value of $x$ where $x+t=1$ (namely, $x=1-t$). The upper integral (for $1-t\le x \le 1)$ is zero because $\mathscr{I}(x+t\lt1)=0$ there, leaving

$$\begin{aligned} &\int_0^1 x(x+t)\mathscr{I}(x+t\lt1)[(x+t)^2\,\mathscr{I}(x+t\lt1)+\mathscr{I}(x+t\ge1)-x^2]^{n-2}\,\mathrm{d}x\,\mathscr{I}(t\gt 0)\\ &=\int_0^{1-t} x(x+t)\left[(x+t)^2-x^2\right]^{n-2}\,\mathrm{d}x. \end{aligned}$$

By the way, the substitution $u = x+t/2$ converts this into the integral

$$(2t)^{n-2}\int_{t/2}^{1-t/2} \left(u^2-\frac{t^2}{4}\right)u^{n-2}\,\mathrm{d}u$$

which has an easy elementary solution.


In general, when indicator functions $\mathscr{I}$ appear in an integrand, they invite you to decompose the integral into the sum of integrals over the regions where the functions constantly equal $0$ or $1.$ On any such region the various indicator functions can be replaced by their constant values, which usually simplify the integrands, reducing the evaluation to that of various simpler integrals. The main complication arises from identifying these regions. In the present case they were formed by the three indicators

$$\mathscr{I}(x+t\lt 1),\ \mathscr{I}(x+t\ge 1),\text{ and } \mathscr{I}(t\gt 0).$$

Given $t\ge 0,$ these determine just two regions of integration, $x\in (-\infty, 1-t)$ and $x\in [1-t,\infty)$ which must be intersected with the domain of integration $x\in[0,1].$

$\endgroup$
0
$\begingroup$

I am not sure if this was the source of your confusion but the notation in the last bit i.e., $1(t>0)$ is an indicator function that takes the value 1 when t is greater than 0 and takes the value 0 otherwise. This only implies that $g(t) = 0$ if $ t \leq 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.