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Let p=1/2 and let B$_n$ = 1 if {w$_n$=H} and B$_n$ = -1 if {w$_n$=T} where w$_n$ represents the outcome of a coin flip at time n.

Given that A$_n$ = B$_n$+B$_{n-1}$+B$_{n-2}$ , calculate E$_3$[A$_4^2$](T,H,H)

My attempt:

A$_4$ = B$_4$+B$_{3}$+B$_{2}$ = B$_4$ + 1 + 1 = B$_4$ + 2

E$_3$[A$_4^2$](T,H,H) = E$_3$[(B$_4$+2)$^2$] =

E$_3$[B$_4^2$+4B$_4$+4]=E$_3$[B$_4^2$]+4E$_3$[B$_4$]+4=

E$_3$[B$_4$]E$_3$[B$_4$] + 4E$_3$[B$_4$] + 4 =

[1(1/2)+(-1)(1/2)][1(1/2)+(-1)(1/2)]+[1(1/2)+(-1)(1/2)]+4 = 0+0+4 = 4

I don't think I can split E$_3$[B$_4^2$] because the expectation of the products of two independent variables is the product of the expectations, but B$_4^2$ is dependent on itself I believe? Any suggestions?

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  • $\begingroup$ What does $\mathbb E_3$ denote? $\endgroup$ – Math1000 Nov 7 '17 at 6:57
  • $\begingroup$ I actually figured it out, I'm suppose to use variance to calculate the squared portion of it instead of separating it into the product of expectations. But it represented the expectation of the variable given information till time n=3 $\endgroup$ – Niko L Nov 13 '17 at 7:09

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