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Stanislav
Stanislav
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If $G(t,x) = f(t) g(x)$, then $$E(\sup(G(t,X))=\sup f(t) E(g(X)1_{g(X)\geq0}) + \inf f(t) E(g(X)1_{g(X)<0})$$ and \begin{equation} \sup E(G(t,X)) = \begin{cases} E(g(X)) \sup f(t) & \text{ if } E(g(X)) \geq 0 \newline E(g(X)) \inf f(t) & \text{ if } E(g(X)) < 0 \end{cases} \end{equation} So the equality holds if $g(X) \geq 0$ a.s. or $g(X) \leq 0$ a.s.

  1. The equality holds if $G(t,x) = f(t) + g(x)$.

  2. If $G(t,x) = f(t) g(x)$, then $$E(\sup(G(t,X))=\sup f(t) E(g(X)1_{g(X)\geq0}) + \inf f(t) E(g(X)1_{g(X)<0})$$ and \begin{equation} \sup E(G(t,X)) = \begin{cases} E(g(X)) \sup f(t) & \text{ if } E(g(X)) \geq 0 \newline E(g(X)) \inf f(t) & \text{ if } E(g(X)) < 0 \end{cases} \end{equation} So the equality holds if $g(X) \geq 0$ a.s. or $g(X) \leq 0$ a.s.

If $G(t,x) = f(t) g(x)$, then $$E(\sup(G(t,X))=\sup f(t) E(g(X)1_{g(X)\geq0}) + \inf f(t) E(g(X)1_{g(X)<0})$$ and \begin{equation} \sup E(G(t,X)) = \begin{cases} E(g(X)) \sup f(t) & \text{ if } E(g(X)) \geq 0 \newline E(g(X)) \inf f(t) & \text{ if } E(g(X)) < 0 \end{cases} \end{equation} So the equality holds if $g(X) \geq 0$ a.s. or $g(X) \leq 0$ a.s.

  1. The equality holds if $G(t,x) = f(t) + g(x)$.

  2. If $G(t,x) = f(t) g(x)$, then $$E(\sup(G(t,X))=\sup f(t) E(g(X)1_{g(X)\geq0}) + \inf f(t) E(g(X)1_{g(X)<0})$$ and \begin{equation} \sup E(G(t,X)) = \begin{cases} E(g(X)) \sup f(t) & \text{ if } E(g(X)) \geq 0 \newline E(g(X)) \inf f(t) & \text{ if } E(g(X)) < 0 \end{cases} \end{equation} So the equality holds if $g(X) \geq 0$ a.s. or $g(X) \leq 0$ a.s.

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Stanislav
Stanislav
  • 401
  • 4
  • 8

If $G(t,x) = f(t) g(x)$, then $$E(\sup(G(t,X))=\sup f(t) E(g(X)1_{g(X)\geq0}) + \inf f(t) E(g(X)1_{g(X)<0})$$ and \begin{equation} \sup E(G(t,X)) = \begin{cases} E(g(X)) \sup f(t) & \text{ if } E(g(X)) \geq 0 \newline E(g(X)) \inf f(t) & \text{ if } E(g(X)) < 0 \end{cases} \end{equation} So the equality holds if $g(X) \geq 0$ a.s. or $g(X) \leq 0$ a.s.