![𝑦𝑖 ො
𝑦𝑖
𝐚𝐫𝐠 𝒎𝒊𝒏
𝜷𝟎, 𝜷𝟏
𝒊=𝟏
𝒏
𝝐𝒊
𝟐
arg min
𝛽0, 𝛽1
𝑖=1
𝑛
[𝑦𝑖 − (𝛽0 + 𝛽1𝑥𝑖)]2
ො
𝑦𝑖 = 𝛽0 + 𝛽1𝑥𝑖
𝑦𝑖 = ො
𝑦𝑖 + 𝜖𝑖
𝝐𝒊 = 𝒚𝒊 − ෝ
𝒚𝒊
𝜷𝟏 =
𝒄𝒐𝒗 𝒙, 𝒚
𝝈𝒙
𝟐
𝜷𝟎 = ഥ
𝒚 − 𝜷𝟏ഥ
𝒙
⇒](https://image.slidesharecdn.com/introductiontomachinelearningwithexamplesinr-220819103316-5f905112/85/Introduction-to-Machine-Learning-with-examples-in-R-26-320.jpg)
Introduction to Machine Learning with examples in R
- 2. 1958
- 3. 1998
- 9. &
- 13. 𝟐 𝜎2 = 1 𝑛 𝑖=1 𝑛 𝑥𝑖 − 𝜇 2 𝜎 = 𝜎2
- 14. 𝜎2 = 4 𝜎 = 2 𝜎2 = 0.57 𝜎 = 0.75
- 16. 𝒙 𝒚 𝒙𝒊 − ഥ 𝒙 𝒚𝒊 − ഥ 𝒚 (𝒙𝒊 − ഥ 𝒙) ⋅ (𝒚𝒊 − ഥ 𝒚) ഥ 𝒙 ഥ 𝒚
- 17. 𝑐𝑜𝑣(𝑋, 𝑌) = σ𝑖=1 𝑛 𝑥𝑖− ҧ 𝑥 ⋅ 𝑦𝑖− ത 𝑦 𝑛
- 19. −1 ≤ 𝑐𝑜𝑣 𝑥, 𝑦 𝜎𝑥 ⋅ 𝜎𝑦 ≤ 1
- 22. 𝑦𝑖 𝑥𝑖 𝑦𝑖 𝑥𝑖 𝑦𝑖 = 𝛽0 + 𝛽1𝑥i + 𝜖𝑖 𝜷𝟎 𝜷𝟏 ො 𝑦𝑖 = 𝛽0 + 𝛽1𝑥i 𝝐𝒊 ො 𝑦𝑖
- 26. 𝑦𝑖 ො 𝑦𝑖 𝐚𝐫𝐠 𝒎𝒊𝒏 𝜷𝟎, 𝜷𝟏 𝒊=𝟏 𝒏 𝝐𝒊 𝟐 arg min 𝛽0, 𝛽1 𝑖=1 𝑛 [𝑦𝑖 − (𝛽0 + 𝛽1𝑥𝑖)]2 ො 𝑦𝑖 = 𝛽0 + 𝛽1𝑥𝑖 𝑦𝑖 = ො 𝑦𝑖 + 𝜖𝑖 𝝐𝒊 = 𝒚𝒊 − ෝ 𝒚𝒊 𝜷𝟏 = 𝒄𝒐𝒗 𝒙, 𝒚 𝝈𝒙 𝟐 𝜷𝟎 = ഥ 𝒚 − 𝜷𝟏ഥ 𝒙 ⇒
- 27. 𝒙 𝒚 𝒙𝒊 − ഥ 𝒙 𝒚𝒊 − ഥ 𝒚 (𝒙𝒊 − ഥ 𝒙) ⋅ (𝒚𝒊 − ഥ 𝒚) 𝒙 − ഥ 𝒙 𝟐 ഥ 𝒙 ഥ 𝒚 𝑟𝑥𝑦 = −125 246 ⋅ 78 = −0,9 𝛽0 = 22 𝛽1 = −0.71 𝜖2 = 20.25 ො 𝑦𝑖 = 22 − 0.71 ⋅ 𝑥𝑖 𝛽1 = 𝑐𝑜𝑣 𝑥, 𝑦 𝜎𝑥 = −125 246 = −0.51 𝜖2 = 14.48 ො 𝑦𝑖 = 19.1 − 0.51 ⋅ 𝑥𝑖 𝛽0 = ത 𝑦 − 𝛽1 ҧ 𝑥 = 14 + 0.51 ⋅ 10 = 19.1
- 28. ෝ 𝒚 = 𝟐𝟓𝟕𝟗𝟐 + 𝟗𝟒𝟓𝟎 ⋅ 𝒀𝒆𝒂𝒓𝒔𝑬𝒙𝒑𝒆𝒓𝒊𝒆𝒏𝒄𝒆
- 30. 𝑖=1 𝑛 𝑦𝑖 − ത 𝑦 2 𝑖=1 𝑛 ො 𝑦𝑖 − ത 𝑦 2 𝑖=1 𝑛 𝑦𝑖 − ො 𝑦 2
- 31. 𝑅2 𝑹𝟐 𝒓𝟐 0 ≤ 𝑅2 ≤ 1
- 35. 𝑥− ҧ 𝑥 𝜎𝑥
- 40. 𝛽1 𝛽1 𝛽1 𝛽1 ≠ 0 𝛽1 𝛽1 ≠ 0 𝑯𝟎: 𝛽1 = 0 𝑯𝟏: 𝛽1 ≠ 0 ො 𝑦𝑖 = 𝛽0 ො 𝑦𝑖 = 𝛽0 + 𝛽1𝑋 𝛽𝑖−0 𝑆𝐸(𝛽𝑖)
- 43. 𝑦 = 𝛽0 + 𝛽1𝑥1 + ⋯ + 𝛽𝑝𝑥𝑝 + 𝜖𝑖 ො 𝑦𝑖 𝛽𝑗 𝑥𝑖
- 46. 𝑯𝟎: 𝛽1 = 𝛽2 = ⋯ = 𝛽𝑝 = 0 𝛽𝑗 ≠ 0 𝑯𝟏: ∃𝑗: 𝛽𝑗≠ 0 ො 𝑦 = 𝛽0 ො 𝑦 = 𝛽0 + 𝛽𝑖𝑋𝑖 + … + 𝛽𝑗𝑋𝑗 ≫ 𝐻0
- 49. 2𝑝 𝟏. 𝟐𝟔𝟕. 𝟔𝟓𝟎. 𝟔𝟎𝟎. 𝟐𝟐𝟖. 𝟐𝟐𝟗. 𝟒𝟎𝟏. 𝟒𝟗𝟔. 𝟕𝟎𝟑. 𝟐𝟎𝟓. 𝟑𝟕𝟔
- 50. 𝑦 = 𝛽0
- 53. 𝑝𝑟𝑖𝑐𝑒 = −22.02 + 0.13 ⋅ 𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 + 0.22 ⋅ 𝑊ℎ𝑒𝑒𝑙𝑏𝑎𝑠𝑒 𝛽1 ⋅ 𝑥1 𝛽2 ⋅ 𝑥2
- 54. 𝑦 = 𝛽0 + 𝛽1𝑥1 + ⋯ + 𝛽𝑝𝑥𝑝
- 55. 𝑝𝑟𝑖𝑐𝑒 = −22.02 + 0.13 ⋅ 𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 + 0.22 ⋅ 𝑊ℎ𝑒𝑒𝑙𝑏𝑎𝑠𝑒 𝛽1 ⋅ 𝑥1 𝛽2 ⋅ 𝑥2
- 58. 𝑥𝑖 = ቊ 1 𝑖𝑓 𝑖 𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑓𝑒𝑚𝑎𝑙𝑒 2 𝑖𝑓 𝑖 𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑚𝑎𝑙𝑒 ො 𝑦𝑖 = ቊ 𝛽0 + 𝛽1 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑓𝑒𝑚𝑎𝑙𝑒 𝛽0 + 2𝛽1 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑚𝑎𝑙𝑒
- 59. 𝑥𝑖1 = ቊ 1 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑎𝑓𝑟𝑖𝑐𝑎𝑛 0 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑓𝑟𝑖𝑐𝑎𝑛 ො 𝑦𝑖 = 𝛽0 + 𝛽1𝑥𝑖1 + 𝛽2𝑥𝑖2 = ቐ 𝜷𝟎 + 𝜷𝟏 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑎𝑓𝑟𝑖𝑐𝑎𝑛 𝜷𝟎 + 𝜷𝟐 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑎𝑠𝑖𝑎𝑡𝑖𝑐 𝜷𝟎 𝑖𝑓 𝑡ℎ𝑒 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑐𝑎𝑢𝑐𝑎𝑠𝑖𝑐 𝑥𝑖2 = ቊ 1 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑎𝑠𝑖𝑎𝑡𝑖𝑐 0 𝑖𝑓 𝑖𝑡ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑠𝑖𝑎𝑡𝑖𝑐
- 61. 𝑦 = 𝛽0 + 𝛽1𝑥1 + 𝛽2𝑥1 𝑛 + ⋯ + 𝜖𝑖
- 68. 𝑝 𝑋 = 𝑒𝛽0+𝛽1𝑋 1 + 𝑒𝛽0+𝛽1𝑋
- 70. 𝑝 𝑋 = 𝑒𝛽0+𝛽1𝑋1+⋯+𝛽𝑝𝑋𝑝 1 + 𝑒𝛽0+𝛽1𝑋1+⋯+𝛽𝑝𝑋𝑝
- 89. መ 𝛿𝑘 𝑥 = 𝑥 ⋅ Ƹ 𝜇𝑘 ො 𝜎2 − Ƹ 𝜇2 2 ො 𝜎2 + 𝑙𝑜𝑔( ො 𝜋𝑘) ෝ 𝝅𝒌
- 103. • • •
- 120. Too many distinctions between observations TOO MANY CLUSTERS
- 121. Find the best balance between information and homogenity of data associated to each cluster