June 2026
Intermediate
384 pages
4h 17m
Japanese
Content preview from ゼロから作るDeep Learning ❻ ―LLM編
![\begin{aligned} \nabla_\theta J(\pi_\theta) &= \mathbb{E}_{\tau \sim \pi_\theta}\left[\sum_{t=0}^T \nabla_\theta \log \pi_\theta(a_t \mid s_t) (R(\tau) - b(s_t))\right] \\ &=\mathbb{E}_{\tau \sim \pi_\theta}\left[\sum_{t=0}^T \nabla_\theta \log \pi_\theta(a_t \mid s_t) R(\tau)\right] \\ & \quad - \mathbb{E}_{\tau \sim \pi_\theta}\left[\sum_{t=0}^T \nabla_\theta \log \pi_\theta(a_t \mid s_t) b(s_t)\right] \end{aligned}](/api/v2/epubs/urn:orm:book:9784814401611/files/images/_review_math/_gen_a3696cbfaf590cabef608e8ca1434728879bfcac023d421bc68124455fb42698.svg)
![\mathbb{E}_{\tau \sim \pi_\theta}\left[\sum_{t=0}^T \nabla_\theta \log \pi_\theta(a_t \mid s_t) b(s_t)\right] = \sum_{t=0}^T \mathbb{E}_{\tau \sim \pi_\theta}\left[\nabla_\theta \log \pi_\theta(a_t \mid s_t) b(s_t)\right]](/api/v2/epubs/urn:orm:book:9784814401611/files/images/_review_math/_gen_68d948c2a5fd7dbf7e0fe818774d14664b88e50f968e4d43db850f207e5699c6.svg)
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付録Cベースライン付き方策勾配の不偏性の証明
「3.7 ベースライン」では、方策勾配法にベースラインを導入しても推定量の不偏性が保たれることを述べました。ここでは、その性質を数学的に証明します。
まず、ベースライン付き方策勾配を2つの項に分解します。
第一項は元の方策勾配です。第二項がベースラインによる追加項であり、この項が0になることを示せば、不偏性が証明できます。
第二項を詳しく見ていきましょう。まず、総和と期待値の順序を入れ替えます。
次に、各時刻
の項について、軌道全体の期待値を「状態
に関する期待値」と「その状態での行動に関する期待値」に分解します。
ここで重要なのは、ベースラインは状態のみに依存し、行動には依存しないという点です。そのため、を行動に関する期待値の外に出すことができます。 ...
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I'm always learning. So when I got on to O'Reilly, I was like a kid in a candy store. There are playlists. There are answers. There's on-demand training. It's worth its weight in gold, in terms of what it allows me to do.