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We know that the empirical risk is : $L_s = \frac{1}{n} \sum_{i=1}^{n} l(f(x_i),z_i)$

where, $n$ = number of samples,$l(f(x),z)$ is a loss function, $S = (z_1,...,z_n)$ are the provided samples to test\train on, $f(x_i)$ is the output of a learning algorithm and $z_i \in Z $ is $(x_i, y_i)$

Some current literature like this (page 3 section 3.2), indicate that the true risk to be the expected value with respect to joint distribution of $n$ samples. $L_\mu = E_{Z^n}[L_s]$ $_{(1)}$

I do not understand how can the true risk be just the expected value of the empirical risk

This is how far I got :

We know that true risk is defined as : $L_u = \int l(f(x),z) dP(x,y) $

Now to write it in a summation form I can assume $n$ to be arbitrarily large, $L_u = \frac{1}{n} \sum_{i=1}^{n} l(f(x_i),z_i).P(x_i,y_i)$ $_{(2)}$

Now have no idea about the next step to reconcile equation $(2)$ and $(1)$

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1 Answer 1

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You don't need to assume $n$ to be arbitrarily large. It's just linearity of expectation:$\DeclareMathOperator{\E}{\mathbb E}$ \begin{align} \E[\text{empirical risk}] &= \E\left[ \frac1n \sum_{i=1}^n l(f(x_i), z_i) \right] \\&= \frac1n \sum_{i=1}^n \E\left[ l(f(x_i), z_i) \right] \\&= \frac1n \sum_{i=1}^n \int l(f(x_i), z_i) \,\mathrm{d}P(x_i, z_i) \\&= \frac1n \sum_{i=1}^n \text{true risk} \\&= \text{true risk} .\end{align}

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  • $\begingroup$ Just to interpret what this means, will the empirical risk always be an unbiased estimator of the true risk ? This seems a little too specific $\endgroup$ Nov 14, 2018 at 18:09
  • $\begingroup$ The empirical risk of a particular model $f$ is an unbiased estimator (assuming iid sampled) of the true risk of that particular estimator, yes. But keep in mind that this will be broken if you've selected $f$ based on those same samples, e.g. by picking the empirical risk minimizer. $\endgroup$
    – Danica
    Nov 14, 2018 at 19:46
  • $\begingroup$ Also if the samples were not iid it would also be a biased estimator probably? Moreover, I did not quite get why using the same samples used for training, not yield the same result? $\endgroup$ Nov 14, 2018 at 20:00
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    $\begingroup$ Using the same samples to train is overfitting. The easiest example to think about might be taking $f$ to be a (1-)nearest-neighbor model, so that if trained on $(x_i, z_i)$, then $f(x_i) = z_i$ exactly for presumably zero loss. Then using the same training set as testing set would give you zero empirical risk, while clearly the true risk (or the empirical risk on a different set) will be nonzero. $\endgroup$
    – Danica
    Nov 14, 2018 at 20:02
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    $\begingroup$ If the samples aren't iid, then it in general won't be an unbiased estimator, but it depends on the particular kind of dependence. $\endgroup$
    – Danica
    Nov 14, 2018 at 20:04

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