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I am given a random variable distributed binomially: $S_2$ ~ $Bin(2, p)$ and the following intervals:

  • $[p_u(0), p_o(0)] = [0, 1 / 4]$
  • $[p_u(1), p_o(1)] = [1 / 4, 3 / 4]$
  • $[p_u(2), p_o(2)] = [1 / 2, 1]$

How can I find $P_p(p < P_u(S_2)), P_p(p > P_o(S_2))$ if

  1. $0 < p < 1 / 4$
  2. $1 / 4 < p < 1 / 2$
  3. $1 / 2 < p < 3 / 4$
  4. $3 / 4 < p < 1$

Any suggestions are highly appreciated!

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  • $\begingroup$ What do things like $p_u(2)$ and $P_p(p < P_u(S_2))$ mean? $\endgroup$
    – Henry
    Sep 14 at 23:26
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$P_u(S_2)$ is a discrete random variable with atoms $\{0,1/4,1/2\}$ and probabilities $[(1-p)^2,2p(1-p),p^2]$. Then

  • $0<p<1/4$) $\qquad\mathbf{P}_p\left(P_u(S_2)>p\right)=\mathbf{P}_p\left(S_2\geq1\right)=2p(1-p)+p^2$
  • $1/4<p<1/2$) $\qquad\mathbf{P}_p\left(P_u(S_2)>p\right)=\mathbf{P}_p\left(S_2=2\right)=p^2$
  • $1/2<p<1$) $\qquad\mathbf{P}_p\left(P_u(S_2)>p\right)=0$.

$P_o(S_2)$ is a discrete RV with atoms $\{1/4,3/4,1\}$ and probabilities $[(1-p)^2,2p(1-p),p^2]$. Then

  • $0<p<1/4$) $\qquad\mathbf{P}_p\left(P_o(S_2)<p\right)=0$
  • $1/4<p<3/4$) $\qquad\mathbf{P}_p\left(P_o(S_2)<p\right)=\mathbf{P}_p\left(S_2=0\right)=(1-p)^2$
  • $3/4<p<1$) $\qquad\mathbf{P}_p\left(P_o(S_2)<p\right)=\mathbf{P}_p\left(S_2\leq1\right)=(1-p)^2+2p(1-p)$.
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