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As an example, suppose $X_{1}, X_{2}, \ldots, X_{n_{1}}$ are iid with pdf $f(x-\theta)$, for $-\infty<$ $x<\infty$, where $-\infty<\theta<\infty$. Denote the cdf of $X_{i}$ by $F(x-\theta) .$ Let $Z_{1}, Z_{2}, \ldots, Z_{n_{2}}$ denote the censored observations. For these observations, we only know that $Z_{j}>a$, for some $a$ that is known, and that the $Z_{j} \mathrm{~s}$ are independent of the $X_{i} \mathrm{~s}$. Then the observed and complete likelihoods are given by $$ \begin{aligned} L(\theta \mid \mathbf{x}) &=[1-F(a-\theta)]^{n_{2}} \prod_{i=1}^{n_{1}} f\left(x_{i}-\theta\right) \\ L^{c}(\theta \mid \mathbf{x}, \mathbf{z}) &=\prod_{i=1}^{n_{1}} f\left(x_{i}-\theta\right) \prod_{i=1}^{n_{2}} f\left(z_{i}-\theta\right) \end{aligned} $$

However, I don't see how the observed likelihood function is derived:

$$ L(\theta \mid \mathbf{x}) =[1-F(a-\theta)]^{n_{2}} \prod_{i=1}^{n_{1}} f\left(x_{i}-\theta\right)$$

In particular where is the component $[1-F(a-\theta)]^{n_{2}}$ coming from?

Ѕᴀᴀᴅ
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ALEXANDER
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1 Answers1

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$X_1,X_2,\dots,X_{n_1},Z_1,Z_2,\dots,Z_{n_2}$ all are iid (indeed, there are a total of $n_1+n_2$ random variables all with the same distribution). For the $X_i$'s, we observe their actual values. But for the $Z_i$'s, we don't actually know the precise values, only that they are greater than a certain constant $a$. This has to be incorporated into the likelihood. That term comes from $\prod_{i=1}^{n_2}P(Z_i>a)=P(Z_1>a)^{n_2}=[1-F(a-\theta)]^{n_2}$.

Loosely speaking the likelihood is $$\begin{split}L(\theta|\textbf x)&=p(X_1=x_1,X_2=x_2,\dots,X_{n_1}=x_{n_1},Z_1>a,Z_2>a,\dots,Z_{n_2}>a)\\ &=p(X_1=x_1)p(X_2=x_2)\dots p(X_{n_1}=x_{n_1})P(Z_1>a)P(Z_2>a)\dots P(Z_{n_2}>a)\\ &=\left(\prod_{i=1}^{n_1}f(x_i|\theta)\right)[1-F(a|\theta)]^{n_2}\end{split}$$

Vons
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  • But if we know that $Z_i$ is larger than a then P($z_i$>a) should that not be 1? As every z would be larger than a. – ALEXANDER Jun 05 '21 at 13:17
  • no thats not right. the definition of the likelihood is the probability of observing the data. to incorporate this information we want P(Zi>a) with Zi the random variable. in fact this is done in practice. if you observe n exponential times but for the n+1st observation you don't wait the full amount of time, but only observe that you have waited at least $a$, then the likelihood is given by $f_n(\textbf x|\beta)P(X>a)=\beta^n e^{-\beta \sum_{i=1}^n x_i}e^{-\beta a}$. hope this sheds light on the matter. – Vons Jun 05 '21 at 16:35
  • So I guess my interpretation is as follows: if I’m integrating the pdf up to a, and subtracting it from 1, then I’m left with the probability of Zi occurring given the parameters. But I guess I want the underlying probability rules that gives me this multiplication. Are you able to add that to your answer? – ALEXANDER Jun 07 '21 at 03:44
  • The probabilities are multiplied together because Zi are independent. Is that what you mean? – Vons Jun 07 '21 at 04:40
  • Yea, there something fundamental that I am missing as I don’t see why they should be multiplied together to give the likelihood of omega given x – ALEXANDER Jun 07 '21 at 04:53
  • If we let Ai be the event that Zi>a, Bcuz the Zi are independent so P(Z1>a,Z2>a,...,Zn2>a)=P(A1,A2,...,An2)=P(A1)P(A2)...P(An2)=P(Z1>a)P(Z2>a)...P(Zn2>a) – Vons Jun 07 '21 at 04:58
  • Should it be $f(x|\theta)$ instead of $f(x-\theta)$? Double check – Vons Jun 07 '21 at 05:02
  • Well, it’s definitely minus and not conditional. And it’s used several places so would hope it’s not a type occurring multiple places in the text. – ALEXANDER Jun 07 '21 at 06:53