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Let π‘‹βˆΌπ–€π—‘π—‰(πœ†), πœ†>0. What is the moment generating function (MGF) 𝑀𝑋(𝑑) of 𝑋?

exp(βˆ’πœ†π‘‘)

πœ†πœ†βˆ’π‘‘βˆ€π‘‘βˆˆβ„

1πœ†

βˆ’πœ†π‘‘βˆ’πœ†βˆ€π‘‘<πœ†

The moment generating function of π‘‹βˆΌπ–€π—‘π—‰(πœ†) is 𝑀𝑋(𝑑) = 1/(1-πœ†π‘‘), 𝑑<1/πœ†.

To see this, we can use the definition of the moment generating function:

𝑀𝑋(𝑑) = 𝔼[𝑒^𝑑𝑋]

= βˆ«β‚€^∞ e^𝑑π‘₯ πœ†e^βˆ’πœ†π‘₯ dx (since 𝑋 has an exponential distribution with parameter πœ†)

= πœ† βˆ«β‚€^∞ e^(𝑑-πœ†)π‘₯ dx

= πœ†/(πœ†-𝑑), 𝑑<πœ†.

Therefore, the moment generating function of 𝑋 is 1/(1-πœ†π‘‘), 𝑑<1/πœ†.

To find the moment generating function (MGF) of π‘‹βˆΌπ–€π—‘π—‰(πœ†), πœ†>0, we can use the formula:

𝑀𝑋(𝑑) = 𝔼[𝑒^(𝑑𝑋)]

In this case, we have π‘‹βˆΌπ–€π—‘π—‰(πœ†), which means 𝑋 is an exponential random variable with rate parameter πœ†. The probability density function (PDF) of 𝑋 is given by:

𝑓(π‘₯) = πœ†π‘’^(βˆ’πœ†π‘₯) for π‘₯ β‰₯ 0, and 𝑓(π‘₯) = 0 for π‘₯ < 0.

To find the MGF, we need to compute the expected value of 𝑒^(𝑑𝑋). Using the definition of expected value:

𝔼[𝑒^(𝑑𝑋)] = ∫[0,∞] 𝑒^(𝑑π‘₯) πœ†π‘’^(βˆ’πœ†π‘₯) 𝑑π‘₯

Simplifying the integrand:

𝔼[𝑒^(𝑑𝑋)] = ∫[0,∞] πœ†π‘’^[(π‘‘βˆ’πœ†)π‘₯] 𝑑π‘₯

Now we can integrate:

𝔼[𝑒^(𝑑𝑋)] = πœ† ∫[0,∞] 𝑒^[(π‘‘βˆ’πœ†)π‘₯] 𝑑π‘₯

Applying the formula for exponentials:

𝔼[𝑒^(𝑑𝑋)] = πœ† (1/(π‘‘βˆ’πœ†)) 𝑒^[(π‘‘βˆ’πœ†)π‘₯] |[0,∞]

Since πœ† > 0 and 𝑑 > πœ†, the limit as π‘₯ approaches infinity is 0. Therefore, the lower bound term contributes:

𝔼[𝑒^(𝑑𝑋)] = πœ† (1/(π‘‘βˆ’πœ†)) 𝑒^[(π‘‘βˆ’πœ†)∞]

Since 𝑒^[(π‘‘βˆ’πœ†)∞] is undefined, this limit does not exist. Hence, the MGF of 𝑋 does not exist.

Therefore, none of the given options are correct.