Find the value of the constant k on a certain pdf

Question

I'm trying to understand a problem that ask for the following:

Given a random variable $X$ that has a pdf $f(x)= k(1-x)^{\frac1\theta-1}$, for x between 0 and 1, find the value of $k$ in function of θ (i'm translating from spanish, I hope it's like this). $\theta$ is a distribution parameter that's always larger than $0$.

I don't understand what the question is asking me to do, ¿could you help me see what I need to do, or maybe show me the first steps?

Answer 1

We use one property of a probability density function. The integral between $0$ and $1$ must be $1$, since $f(x)=0$ for $x<0,x>1$.

See here for the properties of a probability density function.

$$k\cdot \int_0^1 (1-x)^{\frac1\theta-1} \, dx=1$$

We can substitute $\frac1\theta-1$ by constant $c$

$$k\cdot \int_0^1 (1-x)^c \, dx=1$$

And the antiderivative of $(1-x)^c$ is

$-\frac{1}{c+1}\cdot (1-x)^{c+1}$

Can you proceed?

Answer 2

You have to find $k$ such that $f$ is a pdf. In fact, you have to choose $k$ such that $\displaystyle \int_0^1 f(x)dx =1$. Hence, $\displaystyle k = \dfrac{1}{\int_0^1 (1-x)^{\frac{1}{\theta}-1}dx}$.