Let {Xn} be a sequence of random variables such that the following limits (in the regular point-wise sense) hold: and lim E[X₂] = c for some constant c, n→∞ lim Var[X₂] = 0. n→∞ Show that Xn converges to c in the mean-square sense, that is, lim E[(Xn-c)²] = 0. n→∞ Hint: write E[(X - c)²] in a way that it involves Var[Xn].

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Let \(\{X_n\}\) be a sequence of random variables such that the following limits (in the regular point-wise sense) hold:

\[
\lim_{n \to \infty} \mathbb{E}[X_n] = c \quad \text{for some constant } c,
\]

and

\[
\lim_{n \to \infty} \text{Var}[X_n] = 0.
\]

Show that \(X_n\) converges to \(c\) in the mean-square sense, that is,

\[
\lim_{n \to \infty} \mathbb{E}[(X_n - c)^2] = 0.
\]

Hint: write \(\mathbb{E}[(X_n - c)^2]\) in a way that it involves \(\text{Var}[X_n]\).
Transcribed Image Text:Let \(\{X_n\}\) be a sequence of random variables such that the following limits (in the regular point-wise sense) hold: \[ \lim_{n \to \infty} \mathbb{E}[X_n] = c \quad \text{for some constant } c, \] and \[ \lim_{n \to \infty} \text{Var}[X_n] = 0. \] Show that \(X_n\) converges to \(c\) in the mean-square sense, that is, \[ \lim_{n \to \infty} \mathbb{E}[(X_n - c)^2] = 0. \] Hint: write \(\mathbb{E}[(X_n - c)^2]\) in a way that it involves \(\text{Var}[X_n]\).
Expert Solution
Step 1

Given that,

Xn is a sequence of random variables.

limnEXn=c; for some constant c.

limnVarXn=0

Need to prove: limnEXn-c2=0.

 

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