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1

Generally, this is not true: the assumption $o(w^2h^2)$ as $wh\to 0$ does not yield the conclusion. For example, let $k(w)=c/(1+w^4)$ with $c>0$ chosen so that the integral of $k$ is $1$. The function $w^4h^4$ satisfies the assumption $o(w^2h^2)$ as $wh\to 0$, but $$\int k(w) w^4h^4\,dw = \infty \quad \text{ for all } h>0$$ So, you need some ...

0

If $\hat \theta_1$ is a maximum likelihood estimator, then your first method is called a profile likelihood, and is used when you have a model with "nuisance" parameters that you don't care to report. The latter is the maximum of the joint likelihood. Both should give you the same value of $\hat \theta_2$. If $\hat \theta_1$ is some other estimator, then ...

0

Since $R:=\{(p_1,\ldots,p_r):p_1+\cdots+p_r=1\}$ is compact, there is a global maximum of the likelihood function $L:=Ap_1^{X_1}\ldots p_r^{X_r}$. $L$ is non-negative, and is 0 everywhere on the boundary of $R$. Let $L_0$ be the likelihood at the point $\frac1n(X_1,\ldots,X_{r-1})$. If $L$ is 0 everywhere in $R$, then $L_0=0$ is a trivial global maximum. ...

1

It is clear that $X(l) \sim N(\theta, \sigma^2_N)$ since it is the sum of $\theta$ and a mean zero Gaussian RV. If the $N(l)$ are independent then the $X(l)$ are as well, and then the joint distribution of the $X(l)$ is simply the product of the distributions of the individual distributions.

3

As you already know, considering the function $$G_n(x)=\log\left(n+\frac{n-1}{x-1}\right),$$ defined on $(1,+\infty)$, $\hat\lambda_n$ solves the identity $$\hat\lambda_n M_n=G_n(\hat\lambda_n M_n).$$ As you noted, this identity has no analytical solution. However, the function $G_n$ decreases on $(1,+\infty)$ from $G_n(1)=+\infty$ to $G_n(+\infty)=\log(n)$ ...

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