Picture of QuestionSuppose $X$ and $Y$ are independent, Pareto-distributed, with cumulative distributions given by $$Fx(;\lambda)=1 −1/x^\lambda$$ $$Fy(y;\lambda)=1 −1/x^\mu$$
with $x, y\ge 1$ and $\lambda, \mu > 0$. Let $Z = \min\{X, Y\}$ and define the (non)censoring indicator
$$\delta =\begin{cases} 1 & \text{if }X < Y,\\ 0 &\text{otherwise.}\end{cases}$$
How do we Derive the maximum likelihood estimators of $\theta$ and $p$? What function do we create?
Let $Z_1$, ... , $Z_n$ be a random sample from $fZ(z; \theta)$, with $\theta =\lambda + \mu$, and let $\delta_1$, ... , $\delta_n$ be a random sample from $fδ(d; p)$, with $p = \lambda/(\lambda + \mu)$.
I found the frequency function for $\delta$ but I am not sure how to get the density function for $Z$.
Any help is appreciated.
As you found the pmf of $\delta$, I will show how to derive the pdf of $Z$.
\begin{align} F_Z(z) &= P(Z \leq z) \\ &= P(min(X, Y) \leq z) \\ &= 1 - P(min(X, Y) > z) \\ &= 1 - P(X > z, Y > z) \\ &= 1 - P(X > z)P(Y > z) \\ &= 1 - (1-P(X \leq z))(1 - P(Y \leq z)) \\ &= 1 - (1-F_X(z))(1-F_Y(z)) \\ &= F_X(z) + F_Y(z) - F_X(z)F_Y(z) \\ \end{align}
Then, \begin{align} f_Z(z) &= \frac{d}{dz}\{F_X(z) + F_Y(z) - F_X(z)F_Y(z)\} \\ &= f_X(z) + f_Y(z) - f_X(z)F_Y(z) - F_X(z)f_Y(z) \\ \end{align}
We can easily get $f_X(x), f_Y(y)$.
Therefore, \begin{align} f_Z(z) &= f_X(z) + f_Y(z) - f_X(z)F_Y(z) - F_X(z)f_Y(z) \\ &= \lambda z^{-\lambda-1} + \mu z^{-\mu-1} - \lambda z^{-\lambda-1}(1 - z^{-\mu}) - (1 - z^{-\lambda})\mu z^{-\mu-1} \\ &= \lambda z^{-\lambda-1} + \mu z^{-\mu-1} - \lambda z^{-\lambda-1} - \mu z^{-\mu-1} + (\lambda + \mu)z^{-(\lambda+\mu+1)} \\ &= (\lambda + \mu)z^{-(\lambda+\mu+1)} \\ &= \theta z^{-(\theta+1)} \end{align}
Now, you can get the MLE from it.
(I got $\displaystyle \frac{n}{log(\sum z_i)}$. I guess you can make it to the answer.)
I got the pmf of $\delta$ as below.