![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-forum-message-attachments/o/inbox%2F1659719%2F578c40f47a89975e907830ebcb458b62%2FScreenshot%20from%202020-09-15%2014-45-17.png?generation=1600170387288696&alt=media\")
Abstract
\nOne of the most problem of the Kaggle competitions is the uncertainty between CV loss and public/private losses. That's the problem, cause it influences your solution and cause of this problem really good solution might be rejected. This topic doesn't resolve this problem, but it introduces one stand point, that you can use when you decide which solution you have to choose for the final submission.
\nTheory
\nLets say we have some solution S and our aim is to find loss value for every sample, that corresponds S. Lets assume, that each sample has it's own proior loss, that we would get without any:
\nBut in practice after applying these factors we get only posterior losses.
\nTraining & Inference: sample's loss
\nDuring training with fair cross validation we have only noise factors:
\n$$
\nl_{CV}[i] \\sim \\ N(l^0_{CV}[i], \\xi)
\n$$
\nwhere
\n$$
\nl_{CV}[i] - CV \\text{ } posterior \\text{ } sample \\text{ } loss, \\text{ } \\text{ }
\n l_{CV}^0[i] - CV \\text{ } prior \\text{ } sample \\text{ } loss
\n$$
But in the inference time all factors comes out:
\n$$
\nl_{pub}[i] \\sim \\ N(l^0_{pub}[i], \\sigma), \\
\nl_{priv}[i] \\sim \\ N(l^0_{priv}[i], \\sigma),
\n$$
\nwhere
\n$$
\n\\sigma = \\xi + \\sigma^{measure} + \\sigma^{domain}
\n$$
I will call this deviation as domain shifting cause it's a dominant factor: noise level is relatively small and we try to avoid measurement's factor.
\nI use normal distribution cause of it convenient properties and understandable pictures. We can replace normal distribution with anything else and properties, that we're going to use, remain correct with large amount of samples cause of central limit theorem (CLT).
\nTraining & Inference: mean loss
\nNow we can count mean cross validation loss:
\n$$
\nl_{CV, n} = \\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}[i] \\sim \\
\nN(\\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}^0[i], \\frac{\\xi}{\\sqrt{n}}) \\sim \\
\nN(l_{CV, n}^0, \\frac{\\xi}{\\sqrt{n}})
\n$$
And public/private mean losses in the same manner:
\n$$
\nl_{pub, n} = N(l_{pub, n}^0, \\frac{\\sigma}{\\sqrt{n}}), \\
\nl_{priv, n} = N(l_{priv, n}^0, \\frac{\\sigma}{\\sqrt{n}})
\n$$
Now we can assume different parameters and see, what we can get with this simple mathematical model. Finally we will try to count real parameters for this task and turn out what case corresponds this competition.
\nBig data & No domain shifting
\nSimple case: lots of data, no factors except noise:
\n$$
\nn_{CV}, n_{pub}, n_{priv} >> 1, \\
\n\\sigma \\approx \\xi = 0.1
\n$$
Cause of formulas above for mean losses:
\n$$
\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0
\n$$
$$
\nl_{CV, n} \\approx l_{pub, n} \\approx l_{priv, n}
\n$$
Some examples (distributions for mean losses and several samples from these distributions):
\n
What pictures are these?
\nParticular picture corresponds particular solution. Green distribution means private loss, so green crosses in the bottom of the picture means real final competition loss you can get with this solution. The same with orange and blue lines and crosses. If orange cross (public loss) lay left to the green cross (private loss), it means that you private loss overrates your solution and final results will be not so good. Details are in the notebook.
\nBig data & Domain shifting
\nLets add domain shifting:
\n$$
\nn_{CV}, n_{pub}, n_{priv} >> 1, \\
\n\\sigma > \\xi
\n$$
$$
\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0
\n$$
$$
\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}
\n$$
This case completely depends of the level of domain shifting. With relatively large ones we get absolutely random competition (the second picture):
\n![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-forum-message-attachments/o/inbox%2F1659719%2F937ff27f22503be1615c84460e5e82ed%2FScreenshot%20from%202020-09-15%2014-45-26.png?generation=1600170402007405&alt=media\")
![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-forum-message-attachments/o/inbox%2F1659719%2F9deffb4f9b67580a4cd71e9071e9157d%2FScreenshot%20from%202020-09-15%2014-45-36.png?generation=1600170415458845&alt=media\")
Small data & Domain shifting
\nOK, what is the case for this competition? We know all amounts of samples (I've doubled amount for the training set, cause during training you can use several samples for one patients, so lets say you use every patient twice), public/private splitting and only domain shifting remains unknown.
\n$$
\nn_{CV} = 2 \\cdot 176, n_{pub} = 0.15 \\cdot 200, n_{priv} = 0.85 \\cdot 200, \\
\n\\sigma = 0.3, \\xi = 0.1
\n$$
$$
\nl_{pub, n}^0, l_{priv, n}^0 \\neq l_{CV, n}^0
\n$$
$$
\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}
\n$$
There is some suggestion on the pictures above - for details check the notebook with picture's generating. In my opinion, we can get some estimation for domain shifting after submissions, but currently I don't know how to do that.
\n![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-forum-message-attachments/o/inbox%2F1659719%2Fde90b302aeaaf7d1fb69ed51b4db995a%2FScreenshot%20from%202020-09-15%2014-45-45.png?generation=1600170427240593&alt=media\")
Conclusion
\nUnder this theory and under constant for domain shifting we can see on the last picture, that CV loss estimates private loss better, than public loss: there are cases (left lower picture), when orange cross lies so far to the left, that can overrate this solution, but CV losses in this case are next to final private losses. That case may describe what really happen in the leaderboard right now :) But we also can see, that private loss is unstable, so final results might be far enough from any estimations. And the common idea of this topic is getting proof for the suggestion: your idea might be really cool even if it improves only CV without public loss improvement!
", "messages": [ { "id": 1011410, "postDate": "2020-09-15T12:53:56.040Z", "content": "Abstract
\nOne of the most problem of the Kaggle competitions is the uncertainty between CV loss and public/private losses. That's the problem, cause it influences your solution and cause of this problem really good solution might be rejected. This topic doesn't resolve this problem, but it introduces one stand point, that you can use when you decide which solution you have to choose for the final submission.
\nTheory
\nLets say we have some solution S and our aim is to find loss value for every sample, that corresponds S. Lets assume, that each sample has it's own proior loss, that we would get without any:
\nBut in practice after applying these factors we get only posterior losses.
\nTraining & Inference: sample's loss
\nDuring training with fair cross validation we have only noise factors:
\n$$
\nl_{CV}[i] \\sim \\ N(l^0_{CV}[i], \\xi)
\n$$
\nwhere
\n$$
\nl_{CV}[i] - CV \\text{ } posterior \\text{ } sample \\text{ } loss, \\text{ } \\text{ }
\n l_{CV}^0[i] - CV \\text{ } prior \\text{ } sample \\text{ } loss
\n$$
But in the inference time all factors comes out:
\n$$
\nl_{pub}[i] \\sim \\ N(l^0_{pub}[i], \\sigma), \\
\nl_{priv}[i] \\sim \\ N(l^0_{priv}[i], \\sigma),
\n$$
\nwhere
\n$$
\n\\sigma = \\xi + \\sigma^{measure} + \\sigma^{domain}
\n$$
I will call this deviation as domain shifting cause it's a dominant factor: noise level is relatively small and we try to avoid measurement's factor.
\nI use normal distribution cause of it convenient properties and understandable pictures. We can replace normal distribution with anything else and properties, that we're going to use, remain correct with large amount of samples cause of central limit theorem (CLT).
\nTraining & Inference: mean loss
\nNow we can count mean cross validation loss:
\n$$
\nl_{CV, n} = \\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}[i] \\sim \\
\nN(\\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}^0[i], \\frac{\\xi}{\\sqrt{n}}) \\sim \\
\nN(l_{CV, n}^0, \\frac{\\xi}{\\sqrt{n}})
\n$$
And public/private mean losses in the same manner:
\n$$
\nl_{pub, n} = N(l_{pub, n}^0, \\frac{\\sigma}{\\sqrt{n}}), \\
\nl_{priv, n} = N(l_{priv, n}^0, \\frac{\\sigma}{\\sqrt{n}})
\n$$
Now we can assume different parameters and see, what we can get with this simple mathematical model. Finally we will try to count real parameters for this task and turn out what case corresponds this competition.
\nBig data & No domain shifting
\nSimple case: lots of data, no factors except noise:
\n$$
\nn_{CV}, n_{pub}, n_{priv} >> 1, \\
\n\\sigma \\approx \\xi = 0.1
\n$$
Cause of formulas above for mean losses:
\n$$
\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0
\n$$
$$
\nl_{CV, n} \\approx l_{pub, n} \\approx l_{priv, n}
\n$$
Some examples (distributions for mean losses and several samples from these distributions):
\n
What pictures are these?
\nParticular picture corresponds particular solution. Green distribution means private loss, so green crosses in the bottom of the picture means real final competition loss you can get with this solution. The same with orange and blue lines and crosses. If orange cross (public loss) lay left to the green cross (private loss), it means that you private loss overrates your solution and final results will be not so good. Details are in the notebook.
\nBig data & Domain shifting
\nLets add domain shifting:
\n$$
\nn_{CV}, n_{pub}, n_{priv} >> 1, \\
\n\\sigma > \\xi
\n$$
$$
\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0
\n$$
$$
\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}
\n$$
This case completely depends of the level of domain shifting. With relatively large ones we get absolutely random competition (the second picture):
\n
Small data & Domain shifting
\nOK, what is the case for this competition? We know all amounts of samples (I've doubled amount for the training set, cause during training you can use several samples for one patients, so lets say you use every patient twice), public/private splitting and only domain shifting remains unknown.
\n$$
\nn_{CV} = 2 \\cdot 176, n_{pub} = 0.15 \\cdot 200, n_{priv} = 0.85 \\cdot 200, \\
\n\\sigma = 0.3, \\xi = 0.1
\n$$
$$
\nl_{pub, n}^0, l_{priv, n}^0 \\neq l_{CV, n}^0
\n$$
$$
\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}
\n$$
There is some suggestion on the pictures above - for details check the notebook with picture's generating. In my opinion, we can get some estimation for domain shifting after submissions, but currently I don't know how to do that.
\n
Conclusion
\nUnder this theory and under constant for domain shifting we can see on the last picture, that CV loss estimates private loss better, than public loss: there are cases (left lower picture), when orange cross lies so far to the left, that can overrate this solution, but CV losses in this case are next to final private losses. That case may describe what really happen in the leaderboard right now :) But we also can see, that private loss is unstable, so final results might be far enough from any estimations. And the common idea of this topic is getting proof for the suggestion: your idea might be really cool even if it improves only CV without public loss improvement!
", "rawMarkdown": "**Abstract**\n\nOne of the most problem of the Kaggle competitions is the uncertainty between CV loss and public/private losses. That's the problem, cause it influences your solution and cause of this problem really good solution might be rejected. This topic doesn't resolve this problem, but it introduces one stand point, that you can use when you decide which solution you have to choose for the final submission.\n\n**Theory**\n\nLets say we have some solution *S* and our aim is to find loss value for every sample, that corresponds *S*. Lets assume, that each sample has it's own *proior* loss, that we would get without any:\n- noises (both in features and targets)\n- loss measurement techniques (classic CV, splitting by patients, last three measurement, etc.)\n- domain shifting (between train and test)\n\nBut in practice after applying these factors we get only *posterior* losses. \n\n**Training & Inference: sample's loss**\n\nDuring training with fair cross validation we have only noise factors:\n\n$$\nl_{CV}[i] \\sim \\\\ N(l^0_{CV}[i], \\xi)\n$$\nwhere \n$$\nl_{CV}[i] - CV \\text{ } posterior \\text{ } sample \\text{ } loss, \\text{ } \\text{ } \n l_{CV}^0[i] - CV \\text{ } prior \\text{ } sample \\text{ } loss\n$$\n\nBut in the inference time all factors comes out:\n\n$$\nl_{pub}[i] \\sim \\\\ N(l^0_{pub}[i], \\sigma), \\\\\nl_{priv}[i] \\sim \\\\ N(l^0_{priv}[i], \\sigma),\n$$\nwhere\n$$\n\\sigma = \\xi + \\sigma^{measure} + \\sigma^{domain}\n$$\n\nI will call this deviation as *domain shifting* cause it's a dominant factor: noise level is relatively small and we try to avoid measurement's factor.\n\nI use normal distribution cause of it convenient properties and understandable pictures. We can replace normal distribution with anything else and properties, that we're going to use, remain correct with large amount of samples cause of central limit theorem (CLT).\n\n**Training & Inference: mean loss**\n\nNow we can count mean cross validation loss:\n\n$$\nl_{CV, n} = \\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}[i] \\sim \\\\\nN(\\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}^0[i], \\frac{\\xi}{\\sqrt{n}}) \\sim \\\\\nN(l_{CV, n}^0, \\frac{\\xi}{\\sqrt{n}})\n$$\n\nAnd public/private mean losses in the same manner:\n\n$$\nl_{pub, n} = N(l_{pub, n}^0, \\frac{\\sigma}{\\sqrt{n}}), \\\\\nl_{priv, n} = N(l_{priv, n}^0, \\frac{\\sigma}{\\sqrt{n}})\n$$\n\nNow we can assume different parameters and see, what we can get with this simple mathematical model. Finally we will try to count real parameters for this task and turn out what case corresponds this competition. \n\n**Big data & No domain shifting**\n\nSimple case: lots of data, no factors except noise:\n$$\nn_{CV}, n_{pub}, n_{priv} >> 1, \\\\\n\\sigma \\approx \\xi = 0.1\n$$\n\nCause of formulas above for mean losses:\n\n$$\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0\n$$\n\n$$\nl_{CV, n} \\approx l_{pub, n} \\approx l_{priv, n}\n$$\n\nSome examples (distributions for mean losses and several samples from these distributions):\n\n\n\n**What pictures are these?**\n\nParticular picture corresponds particular solution. Green distribution means private loss, so green crosses in the bottom of the picture means real final competition loss you can get with this solution. The same with orange and blue lines and crosses. If orange cross (public loss) lay left to the green cross (private loss), it means that you private loss overrates your solution and final results will be not so good. Details are in the [notebook](https://www.kaggle.com/koza4ukdmitrij/how-to-choose-the-best-solution?scriptVersionId=42728666).\n\n**Big data & Domain shifting**\n\nLets add domain shifting:\n\n$$\nn_{CV}, n_{pub}, n_{priv} >> 1, \\\\\n\\sigma > \\xi\n$$\n\n$$\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0\n$$\n\n$$\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}\n$$\n\nThis case completely depends of the level of domain shifting. With relatively large ones we get absolutely random competition (the second picture):\n\n\n\n\n\n**Small data & Domain shifting**\n\nOK, what is the case for this competition? We know all amounts of samples (I've doubled amount for the training set, cause during training you can use several samples for one patients, so lets say you use every patient twice), public/private splitting and only domain shifting remains unknown.\n\n$$\nn_{CV} = 2 \\cdot 176, n_{pub} = 0.15 \\cdot 200, n_{priv} = 0.85 \\cdot 200, \\\\\n\\sigma = 0.3, \\xi = 0.1\n$$\n\n$$\nl_{pub, n}^0, l_{priv, n}^0 \\neq l_{CV, n}^0\n$$\n\n$$\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}\n$$\n\nThere is some suggestion on the pictures above - for details check the [notebook](https://www.kaggle.com/koza4ukdmitrij/how-to-choose-the-best-solution?scriptVersionId=42728666) with picture's generating. In my opinion, we can get some estimation for domain shifting after submissions, but currently I don't know how to do that.\n\n\n\n\n**Conclusion**\n\nUnder this theory and under constant for domain shifting we can see on the last picture, that CV loss estimates private loss better, than public loss: there are cases (left lower picture), when orange cross lies so far to the left, that can overrate this solution, but CV losses in this case are next to final private losses. That case may describe what really happen in the leaderboard right now :) But we also can see, that private loss is unstable, so final results might be far enough from any estimations. And the common idea of this topic is getting proof for the suggestion: your idea might be really cool even if it improves only CV without public loss improvement!", "votes": 5 } ], "comments": [], "raw_markdown_by_id": { "1011410": "**Abstract**\n\nOne of the most problem of the Kaggle competitions is the uncertainty between CV loss and public/private losses. That's the problem, cause it influences your solution and cause of this problem really good solution might be rejected. This topic doesn't resolve this problem, but it introduces one stand point, that you can use when you decide which solution you have to choose for the final submission.\n\n**Theory**\n\nLets say we have some solution *S* and our aim is to find loss value for every sample, that corresponds *S*. Lets assume, that each sample has it's own *proior* loss, that we would get without any:\n- noises (both in features and targets)\n- loss measurement techniques (classic CV, splitting by patients, last three measurement, etc.)\n- domain shifting (between train and test)\n\nBut in practice after applying these factors we get only *posterior* losses. \n\n**Training & Inference: sample's loss**\n\nDuring training with fair cross validation we have only noise factors:\n\n$$\nl_{CV}[i] \\sim \\\\ N(l^0_{CV}[i], \\xi)\n$$\nwhere \n$$\nl_{CV}[i] - CV \\text{ } posterior \\text{ } sample \\text{ } loss, \\text{ } \\text{ } \n l_{CV}^0[i] - CV \\text{ } prior \\text{ } sample \\text{ } loss\n$$\n\nBut in the inference time all factors comes out:\n\n$$\nl_{pub}[i] \\sim \\\\ N(l^0_{pub}[i], \\sigma), \\\\\nl_{priv}[i] \\sim \\\\ N(l^0_{priv}[i], \\sigma),\n$$\nwhere\n$$\n\\sigma = \\xi + \\sigma^{measure} + \\sigma^{domain}\n$$\n\nI will call this deviation as *domain shifting* cause it's a dominant factor: noise level is relatively small and we try to avoid measurement's factor.\n\nI use normal distribution cause of it convenient properties and understandable pictures. We can replace normal distribution with anything else and properties, that we're going to use, remain correct with large amount of samples cause of central limit theorem (CLT).\n\n**Training & Inference: mean loss**\n\nNow we can count mean cross validation loss:\n\n$$\nl_{CV, n} = \\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}[i] \\sim \\\\\nN(\\frac{1}{n_{CV}}\\sum_{i=1}^{n_{CV}} l_{CV}^0[i], \\frac{\\xi}{\\sqrt{n}}) \\sim \\\\\nN(l_{CV, n}^0, \\frac{\\xi}{\\sqrt{n}})\n$$\n\nAnd public/private mean losses in the same manner:\n\n$$\nl_{pub, n} = N(l_{pub, n}^0, \\frac{\\sigma}{\\sqrt{n}}), \\\\\nl_{priv, n} = N(l_{priv, n}^0, \\frac{\\sigma}{\\sqrt{n}})\n$$\n\nNow we can assume different parameters and see, what we can get with this simple mathematical model. Finally we will try to count real parameters for this task and turn out what case corresponds this competition. \n\n**Big data & No domain shifting**\n\nSimple case: lots of data, no factors except noise:\n$$\nn_{CV}, n_{pub}, n_{priv} >> 1, \\\\\n\\sigma \\approx \\xi = 0.1\n$$\n\nCause of formulas above for mean losses:\n\n$$\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0\n$$\n\n$$\nl_{CV, n} \\approx l_{pub, n} \\approx l_{priv, n}\n$$\n\nSome examples (distributions for mean losses and several samples from these distributions):\n\n\n\n**What pictures are these?**\n\nParticular picture corresponds particular solution. Green distribution means private loss, so green crosses in the bottom of the picture means real final competition loss you can get with this solution. The same with orange and blue lines and crosses. If orange cross (public loss) lay left to the green cross (private loss), it means that you private loss overrates your solution and final results will be not so good. Details are in the [notebook](https://www.kaggle.com/koza4ukdmitrij/how-to-choose-the-best-solution?scriptVersionId=42728666).\n\n**Big data & Domain shifting**\n\nLets add domain shifting:\n\n$$\nn_{CV}, n_{pub}, n_{priv} >> 1, \\\\\n\\sigma > \\xi\n$$\n\n$$\nl_{CV, n}^0 \\approx l_{pub, n}^0 \\approx l_{priv, n}^0\n$$\n\n$$\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}\n$$\n\nThis case completely depends of the level of domain shifting. With relatively large ones we get absolutely random competition (the second picture):\n\n\n\n\n\n**Small data & Domain shifting**\n\nOK, what is the case for this competition? We know all amounts of samples (I've doubled amount for the training set, cause during training you can use several samples for one patients, so lets say you use every patient twice), public/private splitting and only domain shifting remains unknown.\n\n$$\nn_{CV} = 2 \\cdot 176, n_{pub} = 0.15 \\cdot 200, n_{priv} = 0.85 \\cdot 200, \\\\\n\\sigma = 0.3, \\xi = 0.1\n$$\n\n$$\nl_{pub, n}^0, l_{priv, n}^0 \\neq l_{CV, n}^0\n$$\n\n$$\nl_{pub, n}, l_{priv, n} \\neq l_{CV, n}\n$$\n\nThere is some suggestion on the pictures above - for details check the [notebook](https://www.kaggle.com/koza4ukdmitrij/how-to-choose-the-best-solution?scriptVersionId=42728666) with picture's generating. In my opinion, we can get some estimation for domain shifting after submissions, but currently I don't know how to do that.\n\n\n\n\n**Conclusion**\n\nUnder this theory and under constant for domain shifting we can see on the last picture, that CV loss estimates private loss better, than public loss: there are cases (left lower picture), when orange cross lies so far to the left, that can overrate this solution, but CV losses in this case are next to final private losses. That case may describe what really happen in the leaderboard right now :) But we also can see, that private loss is unstable, so final results might be far enough from any estimations. And the common idea of this topic is getting proof for the suggestion: your idea might be really cool even if it improves only CV without public loss improvement!" } }