Calculating Interquartile Range (IQR) for a Continuous Random Variable

The question asks us to find the interquartile range (IQR) for a continuous random variable X whose probability density function (PDF) is given by $f(x) = e^{-x}$ for $x \ge 0$. The IQR is a measure of statistical dispersion, representing the difference between the third quartile ($Q_3$) and the first quartile ($Q_1$).

To find the quartiles $Q_1$ and $Q_3$, we first need to determine the cumulative distribution function (CDF), $F(x)$, of the random variable X. The CDF is defined as $F(x) = P(X \le x)$.

Finding the Cumulative Distribution Function (CDF)

For a continuous random variable with PDF $f(x)$, the CDF $F(x)$ is given by the integral of the PDF from $-\infty$ to $x$.

Given $f(x) = e^{-x}$ for $x \ge 0$ and $f(x) = 0$ for $x < 0$:

• For $x < 0$: $$F(x) = \int_{-\infty}^{x} f(t) dt = \int_{-\infty}^{x} 0 dt = 0$$
• For $x \ge 0$: $$F(x) = \int_{-\infty}^{x} f(t) dt = \int_{-\infty}^{0} 0 dt + \int_{0}^{x} e^{-t} dt$$ $$F(x) = 0 + \left[-e^{-t}\right]_{0}^{x} = -e^{-x} - (-e^{-0}) = -e^{-x} + 1 = 1 - e^{-x}$$

So, the CDF is: $$F(x) = \begin{cases} 0 & \text{for } x < 0 \\ 1 - e^{-x} & \text{for } x \ge 0 \end{cases}$$

Calculating the First Quartile ($Q_1$)

The first quartile, $Q_1$, is the value $x_1$ such that the probability of X being less than or equal to $x_1$ is 0.25 (or 25%). Mathematically, $F(x_1) = 0.25$.

Since $Q_1$ for this distribution is non-negative (as $F(x)=0$ for $x < 0$), we use the form $F(x) = 1 - e^{-x}$: $$1 - e^{-Q_1} = 0.25$$ $$e^{-Q_1} = 1 - 0.25$$ $$e^{-Q_1} = 0.75$$ $$e^{-Q_1} = \frac{3}{4}$$ Taking the natural logarithm of both sides: $$-Q_1 = \ln\left(\frac{3}{4}\right)$$ $$Q_1 = -\ln\left(\frac{3}{4}\right)$$ Using the property $\ln(a/b) = \ln a - \ln b$ and $-\ln(y) = \ln(1/y)$: $$Q_1 = \ln\left(\left(\frac{3}{4}\right)^{-1}\right) = \ln\left(\frac{4}{3}\right)$$

Calculating the Third Quartile ($Q_3$)

The third quartile, $Q_3$, is the value $x_3$ such that the probability of X being less than or equal to $x_3$ is 0.75 (or 75%). Mathematically, $F(x_3) = 0.75$.

Using the form $F(x) = 1 - e^{-x}$ for $x \ge 0$: $$1 - e^{-Q_3} = 0.75$$ $$e^{-Q_3} = 1 - 0.75$$ $$e^{-Q_3} = 0.25$$ $$e^{-Q_3} = \frac{1}{4}$$ Taking the natural logarithm of both sides: $$-Q_3 = \ln\left(\frac{1}{4}\right)$$ $$Q_3 = -\ln\left(\frac{1}{4}\right)$$ Using the property $-\ln(y) = \ln(1/y)$: $$Q_3 = \ln\left(\left(\frac{1}{4}\right)^{-1}\right) = \ln(4)$$

Calculating the Interquartile Range (IQR)

The interquartile range (IQR) is the difference between the third quartile ($Q_3$) and the first quartile ($Q_1$).

$$IQR = Q_3 - Q_1$$ $$IQR = \ln(4) - \ln\left(\frac{4}{3}\right)$$ Using the property $\ln a - \ln b = \ln(a/b)$: $$IQR = \ln\left(\frac{4}{\frac{4}{3}}\right)$$ $$IQR = \ln\left(4 \times \frac{3}{4}\right)$$ $$IQR = \ln(3)$$

Therefore, the interquartile range of the continuous random variable X with the given PDF is $\ln 3$.

Revision Table: Key Concepts for IQR Calculation

Additional Information: Percentiles and Dispersion

Quartiles are specific types of percentiles. The first quartile ($Q_1$) is the 25th percentile, the second quartile ($Q_2$) is the 50th percentile (which is also the median), and the third quartile ($Q_3$) is the 75th percentile.

The IQR is a measure of statistical dispersion or spread. It is a robust measure because it is not affected by extreme outliers, unlike the standard deviation or variance which consider all data points (or the entire distribution for continuous variables).

For a continuous random variable with CDF $F(x)$, the $p$-th percentile ($P_p$) is the value $x_p$ such that $F(x_p) = p/100$. So, $Q_1 = P_{25}$ and $Q_3 = P_{75}$.

In this case, the distribution is an exponential distribution with rate parameter $\lambda = 1$. The CDF of an exponential distribution with parameter $\lambda$ is $F(x) = 1 - e^{-\lambda x}$ for $x \ge 0$. The quartiles can be found by solving $1 - e^{-\lambda Q_p} = p/100$ for $Q_p$.

• For $Q_1$ (25th percentile, $p=25$): $1 - e^{-\lambda Q_1} = 0.25 \implies e^{-\lambda Q_1} = 0.75 \implies -\lambda Q_1 = \ln(0.75) \implies Q_1 = -\frac{1}{\lambda}\ln(0.75) = \frac{1}{\lambda}\ln(4/3)$.
• For $Q_3$ (75th percentile, $p=75$): $1 - e^{-\lambda Q_3} = 0.75 \implies e^{-\lambda Q_3} = 0.25 \implies -\lambda Q_3 = \ln(0.25) \implies Q_3 = -\frac{1}{\lambda}\ln(0.25) = \frac{1}{\lambda}\ln(4)$.

The IQR for an exponential distribution with parameter $\lambda$ is $Q_3 - Q_1 = \frac{1}{\lambda}\ln(4) - \frac{1}{\lambda}\ln(4/3) = \frac{1}{\lambda}(\ln 4 - \ln(4/3)) = \frac{1}{\lambda}\ln(4 / (4/3)) = \frac{1}{\lambda}\ln(3)$.

For this problem, $\lambda=1$, so the IQR is $\frac{1}{1}\ln(3) = \ln 3$. This confirms our earlier calculation.