Order Statistics and Functions of Random Variables

Question

I'd like some help with following problem, please.

$X \sim Bin(5; 0,2) \\ Y = \frac{1}{2}(Max(X,4) + Min(X,2))$

I need to find Y's density, but i don't really know how to apply Jacobian method or something alike here. I thought of doing something like

$Z_{1} = Max(X,4) \quad \quad Z_{2} = Min(X,2) \\ Y =\frac{1}{2}(Z_{1} + Z_{2})$

I know the "regular" order statistics distributions when e.g. Max(A,B), where A and B are both random variables, but i got confused as 4 and 2 are just constants. How should i proceed in the case above? Thanks for the help in advance.

Answer 1

Your question is quite simple. All you need is to compute $Y$ corresponding to $X = 0,1,2,3,4,5$. For example, if $X = 0$ then $Y = \frac{1}{2}(4+0)=2$. Likewise, compute the values of $Y$ for other values of $X$ and add the probabilities to get the probability mass function of $Y$. \begin{array}{|c|c|c|} \hline x & Y & \Pr\{X=x\}\\ \hline 0 & 2 & \binom{5}{0}\times 0.2^0\times0.8^5\\ \hline 1 & 2.5 & \binom{5}{1}\times 0.2^1\times0.8^4\\ \hline 2 & 3 & \binom{5}{2}\times 0.2^2\times0.8^3\\ \hline 3 & 3 & \binom{5}{3}\times 0.2^3\times0.8^2\\ \hline 4 & 3 & \binom{5}{4}\times 0.2^4\times0.8^1\\ \hline 5 & 3.5 & \binom{5}{5}\times 0.2^5\times0.8^0\\ \hline \end{array} Consequently, $\Pr\{Y=2\} = \Pr\{X=0\}=?$, $\Pr\{Y=2.5\} = \Pr\{X=1\}=?$, $\Pr\{Y=3\} = \Pr\{X=2\}+\Pr\{X=3\}+\Pr\{X=4\}=?$, and $\Pr\{Y=3.5\}=\Pr\{X=5\}=?$

Answer 2

Naught to do with order statistics.

$\mathsf P(\max\{X,4\}=k)=\begin{cases}\mathsf P(X=5)&:& k=5\\\mathsf P(X\leq 4)&:& k=4\\ 0 &:& \text{otherwise} \end{cases}\\\mathsf P(\min\{X,2\}=k)=\begin{cases}\mathsf P(X=0)&:& k=0\\\mathsf P(X=1)&:& k=1\\\mathsf P(X\geq 2)&:& k=2\\ 0 &:& \text{otherwise} \end{cases}$

and so forth.