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Question

# Decide whether each function defined as follows is a probability density function for the given interval. $f(x)=\sqrt{x}$; [4, 9]

Solution

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The function $f(x)$ is given to us as :

$f(x) = \sqrt{x}\quad ;\quad x\in [4, 9]$

To check whether it is a pdf or not, we first recall the conditions for a function to be a probability density function.

$\textbf{Conditions for function to be a pdf :}$

The function $f$ is a probability density function of a random variable $X$ in the interval [a, b] if it satisfies following two conditions :

Condition 1 : $f(x)\ge 0$ for all $x$ in the interval [a, b]

Condition 2 : $\int_{a}^{b}f(x)\cdot dx = 1$

As we can see that $f(x)\ge 0$ for all $x$ in the interval $[4, 9]$ , hence it satisfies condition-1

Now we check for condition-2 :

\begin{align*} \int_{4}^{9}f(x)\cdot dx &= \int_{4}^{9}\sqrt{x}\cdot dx\\ &= \left[\dfrac{x^{3/2}}{3/2}\right]_{4}^{9}\qquad\qquad\qquad\qquad\text{Since } \int_{p}^{q} x^{n}\cdot dx = \left[\dfrac{x^{n+1}}{n+1}\right]_{p}^{q}\\ &= \left(\dfrac{2}{3}\right)\left[x^{3/2}\right]_{4}^{9}\\ &= \left(\dfrac{2}{3}\right)\left[9^{3/2} - 4^{3/2}\right]\\ &= \left(\dfrac{2}{3}\right)\left[27 - 8\right]\\ \int_{4}^{9}f(x)\cdot dx &= \dfrac{38}{3} \ne 1 \end{align*}

Since it does not satisfy condition-2, hence $f(x)$ is not a probability density function on the indicated interval.

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