Skip to main content
Tweeted twitter.com/StackStats/status/1219681150974808065
Became Hot Network Question
added 22 characters in body
Source Link
Michael Hardy
  • 11k
  • 1
  • 33
  • 56

Note that $a$ has a mean of 0.

My approach:

$$X_t=X_{t-1}+a_t$$ $$E[X_{t+1}|X_1 + \dots+X_{t-1}]$$$$E[X_{t+1}\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}+2a|X_1 + \dots+X_{t-1}]$$$$=E[X_{t-1}+2a\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]+E[2a|X_1 + \dots+X_{t-1}]$$$$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]+E[2a\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]+0$$$$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]+0$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]$$$$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]$$ $$=X_{t-1}$$ Am I doing something wrong here? shouldn't the end product be $X_{t}$$X_t$?

Note that $a$ has a mean of 0.

My approach:

$$X_t=X_{t-1}+a_t$$ $$E[X_{t+1}|X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}+2a|X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]+E[2a|X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]+0$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]$$ $$=X_{t-1}$$ Am I doing something wrong here? shouldn't the end product be $X_{t}$?

Note that $a$ has a mean of 0.

My approach:

$$X_t=X_{t-1}+a_t$$ $$E[X_{t+1}\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}+2a\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]+E[2a\mid X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]+0$$ $$=E[X_{t-1}\mid X_1 + \dots+X_{t-1}]$$ $$=X_{t-1}$$ Am I doing something wrong here? shouldn't the end product be $X_t$?

Source Link
GarlicSTAT
  • 309
  • 3
  • 9

Prove that a simple random walk is a martingale

Note that $a$ has a mean of 0.

My approach:

$$X_t=X_{t-1}+a_t$$ $$E[X_{t+1}|X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}+2a|X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]+E[2a|X_1 + \dots+X_{t-1}]$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]+0$$ $$=E[X_{t-1}|X_1 + \dots+X_{t-1}]$$ $$=X_{t-1}$$ Am I doing something wrong here? shouldn't the end product be $X_{t}$?