Math Problem Statement
Find the MGF of the random variable X having the pdf οΏ½ οΏ½(π₯) = { π₯, πππ 0<π₯<1 2βπ₯, πππ 1<π₯<2 0, }. οΏ½ οΏ½π‘βπππ€ππ π.
Solution
The moment generating function (MGF) of a random variable is defined as:
where is the probability density function (pdf) of . In this case, we have the following pdf for :
\begin{cases} x & \text{for } 0 < x < 1 \\ 2 - x & \text{for } 1 < x < 2 \\ 0 & \text{otherwise} \end{cases}$$ ### Step 1: Set up the MGF The MGF is computed by integrating $$e^{tx} f_X(x)$$ over the range where $$f_X(x)$$ is non-zero. $$M_X(t) = \int_0^1 e^{tx} x \, dx + \int_1^2 e^{tx} (2 - x) \, dx$$ ### Step 2: Compute the first integral $$I_1 = \int_0^1 e^{tx} x \, dx$$ This is a standard integration problem. We can use integration by parts. Let: $$u = x, \quad dv = e^{tx} dx$$ $$du = dx, \quad v = \frac{e^{tx}}{t}$$ Now apply the integration by parts formula: $$I_1 = \left[ \frac{x e^{tx}}{t} \right]_0^1 - \int_0^1 \frac{e^{tx}}{t} \, dx$$ Evaluating this: $$I_1 = \left( \frac{1 e^{t}}{t} - 0 \right) - \frac{1}{t} \int_0^1 e^{tx} \, dx$$ $$I_1 = \frac{e^t}{t} - \frac{1}{t^2} \left( e^t - 1 \right)$$ So, the first integral is: $$I_1 = \frac{e^t}{t} - \frac{e^t - 1}{t^2}$$ ### Step 3: Compute the second integral $$I_2 = \int_1^2 e^{tx} (2 - x) \, dx$$ This also requires integration by parts. Let: $$u = 2 - x, \quad dv = e^{tx} dx$$ $$du = -dx, \quad v = \frac{e^{tx}}{t}$$ Now apply the integration by parts formula: $$I_2 = \left[ (2 - x) \frac{e^{tx}}{t} \right]_1^2 + \int_1^2 \frac{e^{tx}}{t} \, dx$$ Evaluating the boundary terms: $$I_2 = \left( (2 - 2) \frac{e^{2t}}{t} - (2 - 1) \frac{e^{t}}{t} \right) + \frac{1}{t} \left( e^{2t} - e^t \right)$$ Simplifying this: $$I_2 = - \frac{e^t}{t} + \frac{1}{t} (e^{2t} - e^t)$$ $$I_2 = \frac{e^{2t} - 2e^t}{t}$$ ### Step 4: Combine the results Now, combine the results from the two integrals to find the MGF: $$M_X(t) = I_1 + I_2$$ $$M_X(t) = \left( \frac{e^t}{t} - \frac{e^t - 1}{t^2} \right) + \frac{e^{2t} - 2e^t}{t}$$ Simplifying: $$M_X(t) = \frac{e^t + e^{2t} - 2e^t}{t} - \frac{e^t - 1}{t^2}$$ $$M_X(t) = \frac{e^{2t} - e^t}{t} - \frac{e^t - 1}{t^2}$$ Thus, the moment generating function (MGF) of $$X$$ is: $$M_X(t) = \frac{e^{2t} - e^t}{t} - \frac{e^t - 1}{t^2}$$ --- Would you like further details or clarifications on any part of the solution? Here are some related questions you could explore: 1. How can we use the MGF to find moments of the random variable? 2. What does the MGF tell us about the distribution of $$X$$? 3. How do we compute the moment generating function for different types of distributions (e.g., exponential, normal)? 4. What is the significance of the MGF in probability theory and statistical inference? 5. How can we use the MGF to verify the properties of a distribution (e.g., mean, variance)? **Tip:** The MGF is useful for deriving moments (like the mean and variance) of a distribution directly from the derivative of the MGF at $$t = 0$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Moment Generating Function
Integration
Integration by Parts
Formulas
M_X(t) = E[e^(tX)] = β« e^(tx) f_X(x) dx
Theorems
Moment Generating Function
Integration by Parts
Suitable Grade Level
Undergraduate (College level)
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