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Conditional Variance
Given two random variables, we define the conditional variance of \(X\) given \(Y\) by
\[ \text{Var}(X | Y ) = E(X^2 | Y ) – (\,E(X | Y )\,)^2 \]
Show that
\[ \text{Var}(X ) = E (\text{Var}(X | Y ) ) +\text{Var}( E(X | Y ) ) \,. \]
Of course \( E(X | Y )\) is just random variable so we have that
\[ \text{Var}( E(X | Y ) ) = E [\,E(X | Y )^2\,] – [\,E(\,E(X | Y )\,)\,]^2 \]
also is is useful to recall that \(E(E(X | Y ))= E(X )\).
Random Walk
Let \(\{Z_1, Z_2, \dots, Z_n,\dots\} \) be a sequence of i.i.d random variables such that
\[ P( Z_k = a ) = \begin{cases} \frac14 & \text{ If } a=1 \\\frac14 & \text{ If } a=0\\\frac12 & \text{ If } a=-1 \end{cases}\]
If \(X_{n+1} = X_n + Z_n\) and \(X_0=1\) what is
- \( E( X_{n+1} | X_{n}) \)?
- \( E( X_{n+1}) \)?
- \( \text{Var}( X_{n+1} | X_{n}) \)?
- \( \text{Var}( X_{n+1} )\)?
Notice that \(X_n\) depends only \(Z_{n-1},Z_{n-2},\dots,Z_1\) and hence \(X_n\) is independent of \(Z_n\) !
