Hot answers tagged

4

So, you want $$\int_0^\infty x^2f(x),dx$$ to exist, but $$\int_0^\infty x^3f(x),dx$$ to be infinite. We know that the integral $\int_1^\infty x^{-n}\,dx$ is finite if $n>1$ and infinite if $n\leq 1$. So one possibility is $f(x) \propto 1_{x\geq 1} x^{-4}$. Or, it would look a bit tidier to use $f(x) \propto (1+x)^{-4}$ on $x>0$. Now we need the ...


4

There is no such constant. Density values should always be non-negative. In $[0,1]$, $x-x^2$ is non-negative so $C\geq0$. And, in $[1,2]$, $x-x^2$ is nonpositive so $C\leq0$. The intersection is $C=0$, but then the integral cannot be $1$.


3

You are correct. $p_X(x)=f(x)$ is just another function and $p_X(Y)$ acts like a transformation on the random variable $Y$, thus making it a random variable.


2

Some hints: (b) Find $F_X(x)=P(X\leq x)=P(b/U^{1/a}\leq x)=P(U\geq b^a/x^a)$. Take it from here, find $F_X(x)$, and then differentiate wrt $x$ to find the PDF. (c) Pick some $a,b$ of your choice, and using any programming language you like, create several uniform random variables, for each uniform random, calculate $X=b/U^{1/a}$ and get random $X$'s. Plot ...


1

One of the simplest ways to numerically approximate integral (calculate the "area under the curve") $$ \int_a^b f(x) \, dx $$ is the quadrature, where you approximate the smooth function with a number of rectangular "bins", calculate their areas and sum them up, as shown on the image below taken from the linked Wikipedia article. Now recall that uniform ...


1

The capital letter means the random variable for which the function if the CDF or PDF. If you're just dealing with $X$ and $Y$ as the random variables, it is easy to write $f(x)$ and $g(y)$ and drop the subscript; it is clear from the context that we mean the PDF (ditto for $F(x)$ and $G(y)$ being CDFs) of $X$ and $Y$. When you have many random variables, ...


1

I would suggest, based on the flexibility of the distribution (in shape) and breath in applications, the two-parameter Weibull distribution. Here is a source. [EDIT] Moreover, the source notes that the skewness and coefficient of variation depend only on the shape parameter. A further generalization of the Weibull distribution is the Hyperbolastic ...


Only top voted, non community-wiki answers of a minimum length are eligible