Question

Verify f\left( x \right)=\left\\{ \begin{matrix} 30{{x}^{4}}{{e}^{-6{{x}^{5}}}};x > 0 \\\ 0;otherwise \\\ \end{matrix} \right. for p.d.f. If $f\left( x \right)$ is a p.d.f, then find $P\left( 1 \right)$.

Answer 1

We start solving the problem by recalling the property of the p.d.f (probability distribution function) as $\int\limits_{-\infty }^{\infty }{g\left( x \right)dx}=1$. We then apply this condition for the function $f\left( x \right)$ given in the problem and use the properties of definite integrals to check whether it is p.d.f or not. We then recall the property of probability $P\left( a\le x\le b \right)=\int\limits_{a}^{b}{f\left( x \right)dx}$ and apply this for $P\left( 1 \right)$ to find the required probability.

Complete step-by-step solution:
According to the problem, we are given that to verify whether f\left( x \right)=\left\\{ \begin{matrix} 30{{x}^{4}}{{e}^{-6{{x}^{5}}}};x > 0 \\\ 0;otherwise \\\ \end{matrix} \right. is p.d.f or not and we need to find the value of $P\left( 1 \right)$ if $f\left( x \right)$ is a p.d.f.
We know that the if a given function $g\left( x \right)$ is said to be p.d.f (probability distribution function), then it need to prove the following property: $\int\limits_{-\infty }^{\infty }{g\left( x \right)dx}=1$. Let us verify whether the given function f\left( x \right)=\left\\{ \begin{matrix} 30{{x}^{4}}{{e}^{-6{{x}^{5}}}};x > 0 \\\ 0;otherwise \\\ \end{matrix} \right. satisfies this condition or not.
So, we have $\int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=\int\limits_{-\infty }^{0}{0dx}+\int\limits_{0}^{\infty }{30{{x}^{4}}{{e}^{-6{{x}^{5}}}}dx}$. As we know that for $a < c < b$, we have $\int\limits_{a}^{b}{g\left( x \right)dx}=\int\limits_{a}^{c}{g\left( x \right)dx}+\int\limits_{c}^{b}{g\left( x \right)dx}$.
$\Rightarrow \int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=\int\limits_{0}^{\infty }{30{{x}^{4}}{{e}^{-6{{x}^{5}}}}dx}$ ---(1), as we know that $\int\limits_{a}^{b}{0dx}=0$.
Let us assume $6{{x}^{5}}=t$ ---(2). Let us apply differential on both sides.
So, we get $d\left( 6{{x}^{5}} \right)=d\left( t \right)$.
We know that $d\left( ah\left( x \right) \right)=ad\left( h\left( x \right) \right)$.
$\Rightarrow 6d\left( {{x}^{5}} \right)=dt$.
We know that $d\left( {{x}^{n}} \right)=n{{x}^{n-1}}dx$.
$\Rightarrow 6\times 5\times {{x}^{4}}dx=dt$.
$\Rightarrow 30{{x}^{4}}dx=dt$ ---(3).
Now, let us find the limits of the definite integral in equation (1).
We have upper limit $x=\infty$, which gives $t=6{{\left( \infty \right)}^{5}}= \infty$ as upper limit ---(4).
We have a lower limit $x=0$, which gives $t=6{{\left( 0 \right)}^{5}}=0$ as lower limit ---(5).
Let us substitute equations (2), (3), (4) and (5) in equation (1).
So, we get $\int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=\int\limits_{0}^{\infty }{{{e}^{-t}}dt}$.
We know that $\int{{{e}^{-x}}dx}=-{{e}^{-x}}+C$ and $\int\limits_{a}^{b}{{{f}^{'}}\left( x \right)dx}=\left[ f\left( x \right) \right]_{a}^{b}=f\left( b \right)-f\left( a \right)$.
So, we get $\int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=\left[ -{{e}^{-t}} \right]_{0}^{\infty }$.
$\Rightarrow \int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=\left( -{{e}^{-\infty }} \right)-\left( -{{e}^{0}} \right)$.
$\Rightarrow \int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=\left( -0 \right)-\left( -1 \right)$.
$\Rightarrow \int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=0+1$.
$\Rightarrow \int\limits_{-\infty }^{\infty }{f\left( x \right)dx}=1$. We can see that f\left( x \right)=\left\\{ \begin{matrix} 30{{x}^{4}}{{e}^{-6{{x}^{5}}}};x > 0 \\\ 0;otherwise \\\ \end{matrix} \right. satisfies the condition of p.d.f.
So, the function f\left( x \right)=\left\\{ \begin{matrix} 30{{x}^{4}}{{e}^{-6{{x}^{5}}}};x > 0 \\\ 0;otherwise \\\ \end{matrix} \right. is a p.d.f (probability distribution function).
Now, let us find the value of $P\left( 1 \right)$.
We know that $P\left( a\le x\le b \right)=\int\limits_{a}^{b}{f\left( x \right)dx}$.
So, we get $P\left( 1 \right)=P\left( -1\le x\le 1 \right)=\int\limits_{-1}^{1}{30{{x}^{4}}{{e}^{-6{{x}^{5}}}}dx}$.
We know that $\int\limits_{a}^{a}{g\left( x \right)dx}=0$.
So, we get $P\left( 1 \right)=0$.
We have found the value of $P\left( 1 \right)$ as 0.

Note: We should not substitute $x=1$ in the p.d.f f\left( x \right)=\left\\{ \begin{matrix} 30{{x}^{4}}{{e}^{-6{{x}^{5}}}};x > 0 \\\ 0;otherwise \\\ \end{matrix} \right. to find the value of $P\left( 1 \right)$, as we know that it is an infinitesimal in the given region of ‘x’ and we know that the probability at infinitesimal of continuous function is 0. We should not confuse $P\left( 1 \right)$ with the cumulative distribution function while solving this problem. We should not forget to change the limits of the definite integral while changing the variable inside the integrand. Similarly, we can expect problems to find the c.d.f (cumulative distribution function) of f\left( x \right)=\left\\{ \begin{matrix} 30{{x}^{4}}{{e}^{-6{{x}^{5}}}};x > 0 \\\ 0;otherwise \\\ \end{matrix} \right..