Question

If the probability density function of a continuous random variable X is $f(X=x) = Ke^{-3x}$, $0<x<\infty$, then a solution of $\frac{P^2+1}{P} = \frac{K}{\sqrt{2}}$ is?

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{3}$
• C. $2$
• D. $\frac{1}{\sqrt{3}}$

Hint:

• Normalize the PDF: $\int_{0}^{\infty} Ke^{-3x} dx = 1 \Rightarrow K[-\frac{1}{3}e^{-3x}]_0^\infty = 1 \Rightarrow K/3 = 1 \Rightarrow K=3$.
• Substitute $K=3$ into the equation: $\frac{P^2+1}{P} = \frac{3}{\sqrt{2}}$.
• Rearrange to $\sqrt{2}P^2 - 3P + \sqrt{2} = 0$.
• Solve the quadratic equation for $P$. Solutions are $\sqrt{2}$ and $\frac{1}{\sqrt{2}}$.

Answer 1

The Correct Option is A

For $f(x) = Ke^{-3x}$ to be a PDF for $0<x<\infty$: $\int_{0}^{\infty} Ke^{-3x} dx = 1$.

$K \left[ \frac{e^{-3x}}{-3} \right]_{0}^{\infty} = 1$.

$K \left( \lim_{x \to \infty} \frac{e^{-3x}}{-3} - \frac{e^{0}}{-3} \right) = 1$.

$K \left( 0 - (-\frac{1}{3}) \right) = 1 \Rightarrow K \left(\frac{1}{3}\right) = 1 \Rightarrow K=3$.

The equation becomes $\frac{P^2+1}{P} = \frac{3}{\sqrt{2}}$.

$P + \frac{1}{P} = \frac{3}{\sqrt{2}}$.

Multiply by $\sqrt{2}P$: $\sqrt{2}(P^2+1) = 3P $

$\Rightarrow \sqrt{2}P^2 - 3P + \sqrt{2} = 0$.

Using the quadratic formula $P = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $P = \frac{3 \pm \sqrt{(-3)^2 - 4(\sqrt{2})(\sqrt{2})}}{2\sqrt{2}}$ $P = \frac{3\pm \sqrt{9 - 8}}{2\sqrt{2}} = \frac{3 \pm \sqrt{1}}{2\sqrt{2}} = \frac{3 \pm 1}{2\sqrt{2}}$.

The solutions are $P_1 = \frac{3+1}{2\sqrt{2}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

And $P_2 = \frac{3-1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

Option (a) $\frac{1}{\sqrt{2}}$ is one of these solutions.

\[ \boxed{\frac{1}{\sqrt{2}}} \]