## 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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