Decide whether the functions defined as follows are probability density functions on the indicated intervals. If not, tell why. $f(x)=2 x^{2}$; [-1, 1]

Solutions

VerifiedFrom the definition:

The function $f$ is a probability density function of a random variable $x$ in the interval $[a,b]$ if :

$\begin{align*} &1.f(x)\geq0 \text{ for all x in interval} [a,b]\\ &2.\int_{a}^{b} f(x)\,dx=1. \end{align*}$

We need to check these to conditions.

First condition:

$f(x)=2\cdot x^{2}$

On the interval $[-1,1]$, $x^{2}\geq0$, any number squared is positive number, so this condition is fulfilled.

Second condition:

$\begin{align*} \int_{-1}^{1} 2x^{2}\,dx&=2\int_{-1}^{1} x^2\,dx\\ &=2\frac{x^{2+1}}{2+1}\bigg|_{-1}^{1}\\ &=2\frac{x^{3}}{3}\bigg|_{-1}^{1}\\ &=\frac{2\,x^{3}}{3}\bigg|_{-1}^{}\\ &=\frac{2\cdot 1^{3}}{3}-\frac{2\cdot -1^{3}}{3}\\ &=\frac{2}{3}-\frac{-2}{3}\\ &=\frac{2}{3}+\frac{2}{3}\\ &=\frac{4}{3}\\ \end{align*}$

Second condition is not fulfilled.

The function ƒ is a probability density function if next two condition are satisfied:

$\begin{align*} f\left(x\right)&\geq 0\\ \int_{a}^{b} f\left(x\right)&=1 \end{align*}$

$f\left(x\right)=2x^2$

Function $f$ is increasing, continious and positive function on interval $\left[1,2\right]$. Therefore, that Condition 1 is satisfied.

For Condition 2 check if $\int_{-1}^{1} f\left(x\right)=1$.

$\begin{align*} \int_{-1}^{1} f\left(x\right)&=\int_{-1}^{1}2x^2 dx &&\text{ Use $\int c f\left(x\right) dx=c \int f\left(x\right) dx$}\\ &=2 \int_{-1}^{1}x^2 dx &&\text{$ \int_{a}^{b} x^n dx= \frac{x^{n+1}}{n+1}\Big]_{a}^{b}$}\\ &=2\frac{x^3}{3}\Big]_{-1}^{-1}\\ &=2\left(\frac{1}{3}-\frac{-1}{3}\right)\\ &=2\cdot\frac{2}{3}\\ &=\frac{4}{3}\\ &\ne 1 \end{align*}$

Since Condition 2 is not satisfied, $f\left(x\right)$ is not a probability density function.

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