(Solved): Use MLE to estimate the parameters of the following probability distributions a) Log-normal distrib ...

Expert Answer

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a) Log-typical dissemination (?, ?²):

To gauge the boundaries of the log-typical dissemination utilizing Most extreme Probability Assessment (MLE), we expect we have a bunch of free and indistinguishably circulated perceptions {x?, x?, ..., x?}.

The likelihood thickness capability (PDF) of the log-typical conveyance is given by:

f(x; ?, ?²) = (1/(x??(2?))) * exp(- (ln(x) - ?)²/(2?²))

To track down the MLE assessors for ? and ?², we expand the log-probability capability. Taking the logarithm of the probability capability works on the computations:

log L(?, ?²) = ?[log(f(xi; ?, ?²))]

Separating log L(?, ?²) concerning ? and ?² and setting the subsidiaries to nothing, we can address for the assessors.

Halfway subsidiary as for ?:

? log L(?, ?²)/? ? = ?[(ln(xi) - ?)/?²] = 0

Working on the situation:

?[ln(xi) - ?] = n? - ?[ln(xi)] = 0

n? = ?[ln(xi)]

? = (1/n) * ?[ln(xi)]

Fractional subsidiary regarding ?²:

? log L(?, ?²)/? ?² = ?[((ln(xi) - ?)² - ?²)/(??)] = 0

Working on the situation:

?[(ln(xi) - ?)² - ?²] = ?[(ln(xi) - ?)²] - n?² = 0

?[(ln(xi) - ?)²] = n?²

?² = (1/n) * ?[(ln(xi) - ?)²]

Consequently, the MLE assessors for the log-ordinary dispersion are:

? = (1/n) * ?[ln(xi)]

?² = (1/n) * ?[(ln(xi) - ?)²]

b) Typical dissemination (µ, 5?²):

Likewise, for the ordinary conveyance, we expect we have a bunch of free and indistinguishably circulated perceptions {x?, x?, ..., x?}.

The likelihood thickness capability (PDF) of the typical conveyance is given by:

f(x; µ, ?²) = (1/(?(2??²))) * exp(- (x - µ)²/(2?²))

To track down the MLE assessors for µ and ?², we expand the log-probability capability:

log L(µ, ?²) = ?[log(f(xi; µ, ?²))]

Separating log L(µ, ?²) concerning µ and ?² and setting the subordinates to nothing, we can tackle for the assessors.

Fractional subsidiary regarding µ:

? log L(µ, ?²)/? µ = ?[(xi - µ)/?²] = 0

Working on the situation:

?[xi - µ] = nµ - ?[xi] = 0

nµ = ?[xi]