Find the value of k for a Continuous PDF: f

Math Problem Statement

Let X be a continuous random variable with the following PDF: f_X(x) = { ke^(-x) for x >= 0; 0 otherwise. Find the value of k.

The given problem asks to find the value of $k$ for the probability density function (PDF) $f_X(x)$ of a continuous random variable $X$ .

The PDF is defined as:

k e^{-x} & \text{for } x \geq 0 \\ 0 & \text{otherwise} \end{cases}$$ To find $$k$$, we need to use the property of probability density functions, which states that the total area under the PDF over all possible values of $$x$$ must equal 1. This gives us the following equation: $$\int_{-\infty}^{\infty} f_X(x) \, dx = 1$$ Since $$f_X(x) = 0$$ for $$x < 0$$, the integral reduces to: $$\int_0^\infty k e^{-x} \, dx = 1$$ ### Step-by-step solution: 1. Integrate $$k e^{-x}$$ with respect to $$x$$: $$\int_0^\infty k e^{-x} \, dx = k \int_0^\infty e^{-x} \, dx$$ The integral of $$e^{-x}$$ from 0 to $$\infty$$ is a standard result: $$\int_0^\infty e^{-x} \, dx = 1$$ 2. Therefore, we get: $$k \times 1 = 1$$ 3. Solving for $$k$$: $$k = 1$$ Thus, the value of $$k$$ is $$1$$. Would you like further details or have any questions on this? --- Here are some additional related questions: 1. How would the PDF change if $$f_X(x) = k e^{-2x}$$ for $$x \geq 0$$? 2. What is the cumulative distribution function (CDF) for this random variable? 3. How would you calculate the expected value of $$X$$ given this PDF? 4. Can you compute the variance of $$X$$ using this PDF? 5. How does the PDF of an exponential distribution compare to this one? **Tip:** When working with PDFs, always ensure the integral over the entire range equals 1, as this represents the total probability.

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Probability Density Function (PDF)
Integration

\int_{-\infty}^{\infty} f_X(x) dx = 1
\int_0^\infty e^{-x} dx = 1

PDF normalization theorem

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