A continuous random variable has probability density function p(x) which is proportional to g(x), where: g(x) = 0 if x < -π/2 g(x) = 0 if x > π/2 g(x) = cos(x) otherwise Write c fo... A continuous random variable has probability density function p(x) which is proportional to g(x), where: g(x) = 0 if x < -π/2 g(x) = 0 if x > π/2 g(x) = cos(x) otherwise Write c for the constant of proportionality, so that p(x) = cg(x). (a) What is c? (b) What is P(X > 0)? (c) What is P(|X| <= 1)?
Understand the Problem
The problem defines a probability density function (PDF) p(x) that is proportional to a function g(x). The function g(x) is defined piecewise: it's 0 outside the interval [-π/2, π/2], and cos(x) within that interval. The problem asks us to: (a) Find the constant of proportionality 'c' such that p(x) = c*g(x) is a valid PDF (i.e., integrates to 1 over its entire support). (b) Calculate the probability that the continuous random variable X is greater than 0. (c) Calculate the probability that the absolute value of the continuous random variable X is less than or equal to 1.
Answer
(a) $c = \frac{1}{2}$ (b) $P(X > 0) = \frac{1}{2}$ (c) $P(|X| \le 1) = \sin(1)$
Answer for screen readers
(a) $c = \frac{1}{2}$ (b) $P(X > 0) = \frac{1}{2}$ (c) $P(|X| \le 1) = \sin(1)$
Steps to Solve
- Determine the form of the PDF
Since $p(x) = c \cdot g(x)$ and $g(x) = \cos(x)$ for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ and $g(x) = 0$ otherwise, we have:
$$ p(x) = \begin{cases} c \cos(x) & -\frac{\pi}{2} \le x \le \frac{\pi}{2} \ 0 & \text{otherwise} \end{cases} $$
- Find the constant c
For $p(x)$ to be a valid PDF, it must integrate to 1 over its entire support. That is,
$$ \int_{-\infty}^{\infty} p(x) , dx = 1 $$
Since $p(x)$ is non-zero only on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, this becomes
$$ \int_{-\pi/2}^{\pi/2} c \cos(x) , dx = 1 $$
$$ c \int_{-\pi/2}^{\pi/2} \cos(x) , dx = 1 $$
$$ c [\sin(x)]_{-\pi/2}^{\pi/2} = 1 $$
$$ c \left[ \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right] = 1 $$
$$ c [1 - (-1)] = 1 $$
$$ 2c = 1 $$
$$ c = \frac{1}{2} $$
So, the PDF is $p(x) = \frac{1}{2} \cos(x)$ for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ and $0$ otherwise.
- Calculate $P(X > 0)$
We need to find the probability that $X > 0$. This is given by the integral of the PDF from 0 to $\frac{\pi}{2}$:
$$ P(X > 0) = \int_{0}^{\pi/2} \frac{1}{2} \cos(x) , dx $$
$$ P(X > 0) = \frac{1}{2} \int_{0}^{\pi/2} \cos(x) , dx $$
$$ P(X > 0) = \frac{1}{2} [\sin(x)]_{0}^{\pi/2} $$
$$ P(X > 0) = \frac{1}{2} \left[ \sin\left(\frac{\pi}{2}\right) - \sin(0) \right] $$
$$ P(X > 0) = \frac{1}{2} [1 - 0] $$
$$ P(X > 0) = \frac{1}{2} $$
- Calculate $P(|X| \le 1)$
We need to find the probability that $|X| \le 1$, which means $-1 \le X \le 1$. This is given by the integral of the PDF from -1 to 1:
$$ P(|X| \le 1) = \int_{-1}^{1} \frac{1}{2} \cos(x) , dx $$
$$ P(|X| \le 1) = \frac{1}{2} \int_{-1}^{1} \cos(x) , dx $$
$$ P(|X| \le 1) = \frac{1}{2} [\sin(x)]_{-1}^{1} $$
$$ P(|X| \le 1) = \frac{1}{2} [\sin(1) - \sin(-1)] $$
Since $\sin(-x) = -\sin(x)$, we have:
$$ P(|X| \le 1) = \frac{1}{2} [\sin(1) - (-\sin(1))] $$
$$ P(|X| \le 1) = \frac{1}{2} [2\sin(1)] $$
$$ P(|X| \le 1) = \sin(1) $$
(a) $c = \frac{1}{2}$ (b) $P(X > 0) = \frac{1}{2}$ (c) $P(|X| \le 1) = \sin(1)$
More Information
The value of $\sin(1)$ is approximately 0.84147. This means there is about an 84.15% chance that the absolute value of the random variable X is less than or equal to 1.
Tips
A common mistake is forgetting to normalize the PDF, i.e., not finding the correct value of 'c' such that the integral of the PDF over its entire support equals 1. Another mistake might occur when evaluating the integral, especially with the limits of integration or the trigonometric functions. Also, not remembering that the integral of $cos(x)$ is $sin(x)$ is a common mistake. For part (c), forgetting that $\sin(-x) = -\sin(x)$ would lead to an incorrect answer.
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