Suppose that X1, ... , Xn form a random sample from a distribution for the pdf f(x0) is as follows: 1 f(x|0) = - 1x-01 for 0 < x < 0 2 Also suppose that the value of 0 is unknown (-<< 0). Find the MLE of 0.
The pdf of each observation has the following form:
"f(x|\u03b8) = (\\dfrac{1}{\u03b8}, for 0 \u2264 x \u2264 \u03b8"
"(0, otherwise"
Therefore, the likelihood function has the form
"L(\u03b8) = ( \\dfrac{1}{\u03b8^n} , \\text{ for } 0 \u2264 x_i \u2264 \u03b8 (i = 1, \u00b7 \u00b7 \u00b7 , n)"
"( 0 , otherwise"
It can be seen that the MLE of θ must be a value of θ for which "\u03b8 \u2265 x_i, \\text{ for } i = 1, \u00b7 \u00b7 \u00b7 , n" and
which maximizes "\\dfrac{1}{\u03b8^n}" among all such values. Since "\\dfrac{1}{\u03b8^n}" is a decreasing function of θ, the estimate will be the smallest possible value of θ such that "\u03b8 \u2265 x_i \\text{ for } i = 1, \u00b7 \u00b7 \u00b7 , n" . This value
is "\u03b8 = \\text{ max }(x_1, \u00b7 \u00b7 \u00b7 , x_n)" , it follows that the MLE of θ is "\\hat{\u03b8} = max(X_1, \u00b7 \u00b7 \u00b7 , X_n)"
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