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You can’t tell; the most you can say is that x is between a and d. Also i don’t see how point e is significant. One way to think about what it means geometrically is to take the center of gravity of the points; imagine putting 4 equal sized weights at a b c d on a number line.


Suppose $X_i$ are iid with continuous cdf $F$, $L_n = \min(X_1,\ldots,X_n)$ and $U_n = \max(X_1,\ldots,X_n)$. Then $$\mathbb P(a \le L_n \le U_n \le b) = \mathbb P( X_1,\ldots,X_n \in [a,b]) = (F(b) - F(a))^n$$ and if $F$ corresponds to a pdf $f$, the joint pdf of $(L_n, U_n)$ is $$f_{L_n, U_n}(x,y) = - \dfrac{\partial^2}{\partial x \partial y} (F(y) - F(x)...


Assuming the $X_i$ are i.i.d., $$\begin{aligned} P(X_i-X_{(1)} \leq t) &= P(X_i-t\leq X_{(1)})\\ &= P\left(\bigcap_k X_i-t\leq X_k\right)\\ &= E\left(1_{X_i-t\leq X_1} \ldots 1_{X_i-t\leq X_n}\right)\\ &= \int 1_{x_i-t\leq x_1} \ldots 1_{x_i-t\leq x_n} e^{-x_1}1_{x_1\geq 0}\ldots e^{-x_n}1_{x_n\geq 0} dx_1\ldots dx_n \\ &= \int 1_{-t\...


Instead of $M(\theta),$ which is the product of marginal distributions of the order statistics, it is more logical to work with their joint likelihood, because unlike the observations themselves, the order statistics are not independent. The joint distribution of all the order statistics can be shown to be $$f(x_{(1)},\dots,x_{(n)}|\theta)=n!f_{\theta}(x_{(...


It is not possible to say. It depends on the actual values of $a,b,c,d$. Depending on the dispersion of $a,b,c,d$, the average $(a+b+c+d)/4$ can be anywhere between $a$ and $d$.

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