## Step-by-Step Solution ### **1. Definitions** - **State-Value Function**: \( V^\pi(s) \): The expected return when starting from state \( s \) and following policy \( \pi \). - **Action-Value Function**: \( Q^\pi(s, a) \): The expected return when starting from state \( s \), taking action \( a \), and then following policy \( \pi \). --- ### **2. Conversion Formulas** #### **A. Deriving \( V^\pi(s) \) from \( Q^\pi(s, a) \)** \[ V^\pi(s) = \sum_{a} \pi(a|s) Q^\pi(s, a) \] This is a weighted sum over all possible actions, using the policy's action probabilities. #### **B. Deriving \( Q^\pi(s, a) \) from \( V^\pi(s) \)** \[ Q^\pi(s, a) = \mathbb{E}[R_{t+1} + \gamma V^\pi(S_{t+1}) \mid S_t = s, A_t = a] \] This requires: - Knowing the environment's transition probabilities \( P(s'|s,a) \) - Knowing the reward model \( R(s,a) \) --- ### **3. Computational Implications** #### **A. \( V^\pi(s) \) from \( Q^\pi(s, a) \)** - **Simple**: Just a weighted sum over actions. - **Policy required**: Need to know \( \pi(a|s) \). - **No environment model required**: Does not need transition or reward details. #### **B. \( Q^\pi(s, a) \) from \( V^\pi(s) \)** - **Complex**: Requires summing over next states and expected rewards. - **Environment model required**: Must know \( P(s'|s,a) \) and \( R(s,a) \). - **More computation**: For each \((s, a)\), you must compute an expected value over all possible next states. --- ### **4. Final Answer** **It is *not easy* to derive \( Q^\pi(s, a) \) from \( V^\pi(s) \).** #### **Justification:** - To compute \( Q^\pi(s, a) \) from \( V^\pi(s) \), you need the full model of the environment (transition and reward functions). - In contrast, deriving \( V^\pi(s) \) from \( Q^\pi(s, a) \) only needs the policy and is a simple calculation. - Therefore, **deriving \( Q^\pi(s, a) \) from \( V^\pi(s) \) is the harder conversion**. --- ## **Summary Table** | Conversion | Model Required | Computational Cost | Difficulty | |---------------------------------------------------|:--------------:|:-----------------:|:----------:| | \( V^\pi(s) \) from \( Q^\pi(s, a) \) | No | Low | Easy | | \( Q^\pi(s, a) \) from \( V^\pi(s) \) | Yes | High | Hard | --- ## **Final Answer (in words)** > Deriving \( Q^\pi(s, a) \) from \( V^\pi(s) \) is not easy because it requires a full model of the environment to compute expected rewards and transitions, whereas deriving \( V^\pi(s) \) from \( Q^\pi(s, a) \) is a simple weighted sum using the policy. Thus, the conversion from \( V^\pi(s) \) to \( Q^\pi(s, a) \) is computationally more demanding.