Conditional Probability – Problem (19/365)

A problem asks the following. Let \phi, \theta_1, \dots, \theta_n be independent random variables where \{ \theta_i \} are identically distributed and \phi takes values 1, \dots, n. Show that if S_\phi = \theta_1 + \dots + \theta_\phi then

\displaystyle  E(S_\phi | \phi) = \phi E\theta_1 \\ V(S_\phi | \phi) = \phi V\theta_1 \\ E S_\phi = E \phi E \theta_1 \\ V S_\phi = E \phi V \theta_1 + V \phi (E \theta_1)^2

The first two follow due to \{ \theta_i \} being idependenty and identically distributed

\displaystyle  E(S_\phi | \phi) = E \theta_1 + E \theta_2 + \dots + E \theta_\phi = \phi E \theta_1 \\ V(S_\phi | \phi) = V S_1 + V S_2 + \dots + V S_\phi = \phi V \theta_1

The last two we compute by using 1) the generalized total probability formula as seen here and 2) the conditional variance formula as seen here.

\displaystyle  E S_\phi = E E(S_\phi | \phi) = E \phi E \theta_1 \\ V S_\phi = EV(S_\phi | \phi) + VE(S_\phi | \phi) \\ = E \phi V \theta_1 + V(\phi E \theta_1) \\ = E \phi V \theta_1 + \phi (E \theta_1)^2 \text{ because } V(a \theta + b) = a^2 V \theta

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