Please see the image below. Let \( X_1, \ldots, X_n \) be iid pdf\[f(x) = \frac{1}{x \sqrt{2\pi \theta_2}} e^{-\frac{(\log(x) - \theta_1)^2}{2\theta_2}}, \quad -\infty < x < \infty,\]and unknown parameters \( \theta_1 \) and \( \theta_2 \). Find the maximum likelihood estimators for \( \theta_1 \) and \( \theta_2 \), respectively. You do NOT have to do a second derivative test for this exercise.

The given density function is: fx=1x2πθ2e-log x-θ122θ2 -∞<x<∞The likelihood function can be written as: L≡Lx, θ=∏ifxi =∏i1xi2πθ2e-log xi-θ122θ2 =2πθ2n2×∏i1xi×e-∑ilog xi-θ12θ22 Taking log both side, log L≡log Lx, θ =n2×log 2πθ2+∑log i1xi-∑ilog xi-θ122θ2 The MLE for the parameter θ1 can be determined as: From the above step, the likelihood function for θ1 is, log L1≡log Lx, θ1=n2×log 2πθ2+∑log i1xi-12θ2∑ilog xi-θ12 differentiating with respect parameter, θ1, ∂log L1∂ θ1=-12θ2∑ilog xi-θ10-1=12θ2∑ilog xi-θ1 Further, ∂log L1∂ θ1=012θ2∑ilog xi-θ1=0∑ilog xi-nθ12θ2=0θ1=∑ilog xinFurther, ∂2log L1∂ θ21<0∂2log L1∂ θ21θ^1=∑ilog xin<0 Thus, the MLE for the parameter θ1 is, MLE θ^1=∑ilog xinThe MLE for the parameter θ2 can be determined as: From step 2, the likelihood function for θ2 is, log L2≡log Lx, θ2=n2×log 2πθ2+∑log i1xi-12θ2∑ilog xi-θ12 differentiating with respect parameter, θ2, ∂log L1∂ θ2=n2×1θ2-12θ22∑ilog xi-θ12 Further, ∂log L2∂ θ2=0n2×1θ2-12θ22∑ilog xi-θ12=0n2=12θ2∑ilog xi-θ12θ2=1n∑ilog xi-θ12Further, ∂2log L2∂ θ22<0∂2log L2∂ θ22/θ^2=∑ilog xi-θ^1n<0 Thus, the MLE for the parameter θ2 is, MLE θ^2=∑ilog xi- θ^12n