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Assume the lifetime (in months) of batteries is modeled as an exponential random variable with parameter ๐œ† . The probability density function of the random variable is ๐‘ ๐œ† ( ๐‘ฅ ) = ๐œ† ๐‘’ โˆ’ ๐œ† ๐‘ฅ , with ๐‘ฅ โˆˆ [ 0 , โˆž ) . We have collected an i.i.d. sample of 3 batteries whose lifetimes in months are 8, 10, 9, respectively. The standardized log-likelihood of a sample with ๐‘› i.i.d. observations is 1 ๐‘› log ๐ฟ ( ๐‘ฅ , ๐œ† ) = 1 ๐‘› โˆ‘ ๐‘– = 1 ๐‘› ( log ( ๐œ† ) โˆ’ ๐œ† ๐‘ฅ ๐‘– ) . Given this information what is the Maximum Likelihood estimate of ๐œ† ?

Options
A.๐œ† ฬ‚ = 1 9 .
B.๐œ† ฬ‚ = 9 .
C.๐œ† ฬ‚ = log ( 9 ) .
D.There is not enough information to compute the estimate of ๐œ† .
E.๐œ† ฬ‚ = 1 log ( 9 ) .
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Step-by-Step Analysis
We start by identifying the model and the data given. The lifetimes of batteries follow an exponential distribution with parameter lambda, and we have an i.i.d. sample of lifetimes: 8, 10, and 9 months, with n = 3. The standardized log-likelihood provided is (1/n) โˆ‘ (log(lambda) โˆ’ lambda x_i) = log(lambda) โˆ’ lambda xฬ„, where xฬ„ is the sample mean. F......Login to view full explanation

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