针对性的公式类似RMSE测量气象模型输出的变化迫使影响-地球科学堆栈交换江南电子竞技平台江南体育网页版 最近30从www.hoelymoley.com 2023 - 07 - 10 - t05:58:32z //www.hoelymoley.com/feeds/question/22558 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/22558 2 针对性的公式类似RMSE测量气象迫使对模型输出的变化的影响 AnthonyB //www.hoelymoley.com/users/22090 2021 - 07 - 18 - t11:35:50z 2021 - 07 - 22 - t08:46:31z < p >我想验证和比较几种参数化模型。我有一些观测计算每个参数化的RMSE最好的气象强迫。但是我想细微的变化迫使。我认为更多的强迫影响输出范围的不确定性,参数化解释RMSE的改进。< / p > < p >在我的例子中,我有5个迫使文件和两个不同的参数化,输出类< span = " math-container " > Y < / span >美元是一个时间序列。我名字< span class = " math-container " >美元Y_ {ij} (t) < / span >美元每个输出与迫使< span class = " math-container " >我< / span >美元和参数<跨类= " math-container " > $ j $ < / span >,我名字的意思是< span class = " math-container " >美元Y_ {ij} (t) < / span >美元为每个时间步之间所有的强迫与参数化j < span class = " math-container " > \酒吧{Y_j} (t)美元< / span >的时间序列观察是<跨类= " math-container " > $ y (t) $ < / span > < / p > < p >我想比较RMSE用以下公式(在那里我替换< span class = " math-container " > $ y (t) $ < / span >, <跨类= " math-container " > \酒吧{Y_j} (t)美元< / span >): < / p > < p > <跨类= " math-container " > $ $ \压裂{1}{N_t} \ sum_k (Y_ {ij} {Y_j} (t_k) - \酒吧(t_k)) ^ 2 $ $ < / span > < / p > < p >最后,我想总结,如果比值<跨类= " math-container " > \美元压裂{RMSE}{\√6{\压裂{1}{N_t} \ sum_k (Y_ {ij} {Y_j} (t_k) - \酒吧(t_k)) ^ 2}} $ < / span >是接近1所以很难断定因为迫使上的错误导致最敏感的输出,如果它是接近0,参数化真的提高了模型适合观测。< / p > < p >是这种方法有关吗? < / p > //www.hoelymoley.com/questions/22558/-/22579 # 22579 2 答案由BarocliniCplusplus针对性的公式类似RMSE测量气象迫使对模型输出的变化的影响 BarocliniCplusplus //www.hoelymoley.com/users/704 2021 - 07 - 21 - t19:39:51z 2021 - 07 - 21 - t19:39:51z < p >我认为你可能会搞混了。让我们按照标准通用的< a href = " https://en.wikipedia.org/wiki/Ensemble_Kalman_filter # Kalman_filter”rel = " nofollow noreferrer " >数据同化< / >并调用您的模型的输出(<跨类= " math-container " > $ k ^{\文本{th}} < / span >参数化/迫使美元)< span class = " math-container " > xk < / span >美元和观察类< span = " math-container " > y < / span >美元。为简单起见,假设观测算子是酉(<跨类= " math-container " > H =我美元< / span >)。方程根均方误差(RMSE) < span class = " math-container " > $ k ^{\文本{th}} $ < / span >模型(<跨类= " math-container " > k < / span >美元可能是唯一索引,迫使和参数化)是类< span = " math-container " > $ $ RMSE_k = \√6 {\ sum_i \压裂{\离开(间{i、k} -y_i \右)^ 2}{N}} $ $ < / span >, < span class = " math-container " > x_i美元< / span >和<跨类= " math-container " > y_i < / span >是美元指数成对“样本”<跨类= " math-container " > X美元< / span >和<跨类= " math-container " > Y < / span >美元。采样使用执行RMSE采样随着时间的推移,但是你可以很容易样本在空间。< / p > < p >什么是度量,很容易与RMSE度量模型变化吗?让我们按照你建议(的故事)。如果我们让意味着国家< span class = " math-container " > $ k ^{\文本{th}} $ < / span >模型类< span = " math-container " > $ {x} _k \酒吧= \压裂{1}{N} \ sum_i间{i、k} $ < / span >那么你建议的公式是<跨类= " math-container " > $ $ \√6 {\ sum_i \压裂{\离开(间{i、k} - {x} \酒吧_k \右)^ 2}{N}} $ $ < / span >。现在看起来很像的公式< a href = " https://en.wikipedia.org/wiki/Standard_deviation " rel = " nofollow noreferrer " >标准差< / >。相反,我们建议这个公式显示依赖于模式的标准偏差波动。然而,如果我们做一个模型意味着变量类< span = " math-container " > $ \帽子{x} _i = \压裂{1}{N} \ sum_k间{i、k} < / span >美元,那么我们就可以确定有多少不同的营力和参数化导致输出随时间:< span class = " math-container " > $ $ \ sigma_i = \√6 {\ sum_k \压裂{\离开(间{i、k} - {x} _i \ \帽子右)^ 2}{N}} $ $ < / span >。< / p > < p >所以真的,你的公式应该是什么样的(恢复回到你的符号):< span class = " math-container " > $ $ \压裂{RMSE}{\√6{\压裂{1}{N} (Y_ {i, j} {Y_j} (t_k) - \酒吧(t_k)) ^ 2}} $ $ < / span > < / p > < p >现在,我认为你的想法很好,但是解释是不正确的。 You can legitimately have a number that is greater than one (for example, if you see no difference, the standard deviation is 0, therefore your metric reaches infinity). You also cannot say if the inclusion of the parameterization makes the model better based on this metric. That would require the examining the RMSE of each parameterization + forcing. Such an experiment shows relative roles that the parameterizations/forcings of similar kind have in creating the number of possible model outputs that could be causing model errors.

An example that I know of where such an analysis was conducted was Thomas et al. (2019). In it, the RMSE was computed and the model standard deviation was compared, with the standard deviation being smaller than the RMSE (therefore leading to numbers greater than 1, per the corrections [namely the square root] to your logic).

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