Let ๐ be a random variable exponentially distributed with parameter ๐ > 0.โฆ
2024
Let ๐ be a random variable exponentially distributed with parameter ๐ > 0. The probability density function of X is given by:
\(f_X(x) = \begin{cases} \lambda e^{-\lambda x}, & \text{if } x \geq 0 \\ 0, & \text{otherwise} \end{cases} \)
If 5๐ธ(๐) = ๐๐๐(๐), where ๐ธ(๐) and ๐๐๐(๐) indicate the expectation and variance of ๐, respectively, the value of ๐ is ______ (rounded off to one decimal place).
Answer: 0.2 โ Answer: 0.2 (rounded to one decimal place) For an exponential distribution with parameter ฮป: Expectation E(X) = 1/ฮป and variance Var(X) = 1/ฮปยฒ. Use the givenโฆ
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Correct answer: 0.2
Answer: 0.2 (rounded to one decimal place)
For an exponential distribution with parameter ฮป:
Expectation E(X) = 1/ฮป and variance Var(X) = 1/ฮปยฒ.
Use the given relation 5 E(X) = Var(X), so 5(1/ฮป) = 1/ฮปยฒ.
Solve the equation: 5/ฮป = 1/ฮปยฒ โ multiply both sides by ฮปยฒ to get 5ฮป = 1.
Therefore ฮป = 1/5 = 0.2, which is positive and satisfies the parameter constraint.