I am looking for an intuitive explanation of the bias-variance tradeoff, both in general and specifically in the context of linear regression.


Imagine some 2D data--let's say height versus weight for students at a high school--plotted on a pair of axes.

Now suppose you fit a straight line through it. This line, which of course represents a set of predicted values, has zero statistical variance. But the bias is (probably) high--i.e., it doesn't fit the data very well.

Next, suppose you model the data with a high-degree polynomial spline. You're not satisfied with the fit, so you increase the polynomial degree until the fit improves (and it will, to arbitrary precision, in fact). Now you have a situation with bias that tends to zero, but the variance is very high.

Note that the bias-variance trade-off doesn't describe a proportional relationship--i.e., if you plot bias versus variance you won't necessarily see a straight line through the origin with slope -1. In the polynomial spline example above, reducing the degree almost certainly increases the variance much less than it decreases the bias.

The bias-variance tradeoff is also embedded in the sum-of-squares error function. Below, I have rewritten (but not altered) the usual form of this equation to emphasize this:

alt text

On the right-hand side, there are three terms: the first of these is just the irreducible error (the variance in the data itself); this is beyond our control so ignore it. The second term is the square of the bias; and the third is the variance. It's easy to see that as one goes up the other goes down--they can't both vary together in the same direction. Put another way, you can think of least-squares regression as (implicitly) finding the optimal combination of bias and variance from among candidate models.

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    $\begingroup$ I'm having trouble understanding the equation; I can't find any way to justify it. Even a basic units analysis identifies problems. Suppose $y$ is measured in parsecs and $x$ in drams, for example, so that $f$ and its estimator (is that what the little dot over the $f$ means?) are also in parsecs. Then the lhs and $\sigma^2$ are squared parsecs; the middle term on the rhs is the square of a difference between a parsec ($f(x)$) and parsecs per dram (due to the division by $x$); and the last term on the rhs is squared parsecs per dram. It's not valid to add any of these terms to one another! $\endgroup$ – whuber Nov 7 '10 at 21:58
  • $\begingroup$ the equation's fine (the little greek letters in the numerator are not 'x' but 'kappa'). Try this: begin w/ a formula for SSE that you are comfortable with and just a few steps will get you to the one above. $\endgroup$ – doug Nov 8 '10 at 0:49
  • $\begingroup$ What is 'kappa' in this context? $\endgroup$ – user28 Nov 9 '10 at 12:12
  • $\begingroup$ I am a noob. Can you help me intuitively see why in the first part of your answer you say that fitting a polynomial spline causes the variance to increase? $\endgroup$ – Rohit Banga Oct 6 '11 at 3:01
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    $\begingroup$ A simpler example: y=a+bx+e(rror). If I fit a constant to this, bias = bx and variance = var(e) + the variance of my estimate of a around the true value. If I add a term b*x to the model, bias is zero everywhere, but now variance includes the effects of the error of my estimate of b as well as of a and the variance of e, so will be higher than in the first case. So there's a tradeoff between decreased bias, obtained by adding terms to the model that ought to be there, and increased variance, obtained by having to estimate those terms and, possibly, adding irrelevant ones. $\endgroup$ – jbowman Nov 29 '11 at 23:47

Let's say you are considering catastrophic health insurance, and there is a 1% probability of getting sick which would cost 1 million dollars. The expected cost of getting sick is thus 10,000 dollars. The insurance company, wanting to make a profit, will charge you 15,000 for the policy.

Buying the policy gives an expected cost to you of 15,000, which has a variance of 0 but can be thought of as biased since it is 5,000 more than the real expected cost of getting sick.

Not buying the policy gives an expected cost of 10,000, which is unbiased since it is equal to the true expected cost of getting sick, but has a very high variance. The tradeoff here is between an approach that is consistently wrong but never by much and an approach that is correct on average but is more variable.


I highly recommend having a look at Caltech ML course by Yaser Abu-Mostafa, Lecture 8 (Bias-Variance Tradeoff) . Here are the outlines:

Say you are trying to learn the sine function:

alt text

Our training set consists of only 2 data points.

Let's try to do it with two models, $h_0(x)=b$ and $h_1(x)=ax+b$:

For $h_0(x)=b$, when we try with many different training sets (i.e. we repeatedly select 2 data points and perform the learning on them), we obtain (left graph represents all the learnt models, right graph represent their mean g and their variance (grey area)):

enter image description here

For $h_1(x)=ax+b$, when we try with many different training sets, we obtain:

enter image description here

If we compare the learnt model with $h_0$ and $h_1$, we can see that $h_0$ yields more simple models than $h_1$, hence a lower variance when we consider all the models learnt with $h_0$, but the best model g (in red on the graph) learnt with $h_1$ is better than the best model learnt g with $h_0$, hence a lower bias with $h_1$:

enter image description here

If you look at the evolution of the cost function with respect to the size of the training set (figures from Coursera - Machine Learning by Andrew Ng):

High bias:

enter image description here

High variance:

enter image description here

  • $\begingroup$ +1, very elaborative. Take $h_1$ as the example, We have drawn many different samples $D_i$, each with 2 points, so we trained many models each with different model parameter estimate $(\hat a_i,\hat b_i)$, right? And for a specific input $x_0$, different $(\hat a_i,\hat b_i)$ gives different prediction $\hat y_0$, so the predicted target value $\hat y_0$ is a random variable varying with the model parameter estimate $(\hat a,\hat b)$. And the bias and variance in your fig are computed for each $x$'s predicted $\hat y$, right? Am I correct? $\endgroup$ – avocado Dec 27 '13 at 9:40
  • $\begingroup$ your sine function is inverted X-P $\endgroup$ – Diego Mar 1 '15 at 19:16
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    $\begingroup$ This is confusing b/c you seem to be referring to plots that aren't there (maybe the "alt text"s). $\endgroup$ – gung Jan 17 '17 at 21:54
  • $\begingroup$ @gung fixed, thanks for pointing it out. ImageShack shut down free accounts and deleting images. And Stack Exchange failed to take appropriate measures. Related: Ban ImageShack images because they are reusing old URLs for advertising (please support the ban) ; What is the easiest way for me to download all my questions+answers across all Stack Exchange sites? (I am glad I had a backup; please push StackExchange to provide better tools to back up content) $\endgroup$ – Franck Dernoncourt Jan 18 '17 at 18:24
  • $\begingroup$ Thanks for sharing the videos link, It explain well what I was looking for, can now understand your answer $\endgroup$ – Espoir Murhabazi Jul 2 '18 at 9:37

I usually think of these two pictures:

First, meaning of bias and variance:

Understanding bias and Variance

Imagine the center of the Red bulls' eye region is the true mean value of our target random variable which we are trying to predict, and the red region indicates the variance spread of this variable. Every time we take a sample set of observations and predict the value of this variable we plot a blue dot. We predicted correctly if the blue dot falls inside the red region. In other words, bias is the measure of how far off are the predicted blue dots from the true red region, intuitively this is an error. Variance is how scattered are our predictions.

Now the trade-off here is:

The trade-off between Bias and Variance

when we try to decrease one of this parameter (either bias or the variance), the other parameter increases. But there is a sweet spot somewhere in between not-so-less bias and not-so-less variance which produces least prediction error in the long run.

These pictures are taken from http://scott.fortmann-roe.com/docs/BiasVariance.html . Checkout the explanations with linear regression and K-nearest neighbors for more details

  • $\begingroup$ the first figure looks more like precision vs accuracy? $\endgroup$ – KingBoomie Nov 29 '16 at 10:59

Here is a very simple explanation. Imagine you have a scatter plot of points {x_i,y_i} that were sampled from some distribution. You want to fit some model to it. You can choose a linear curve or a higher order polynomial curve or something else. Whatever you choose is going to be applied to predict new y values for a set of {x_i} points. Let's call these the validation set. Let's assume that you also know their true {y_i} values and we are using these just to test out model.

The predicted values are going to be different from the real values. We can measure the properties of their differences. Let's just consider a single validation point. Call it x_v and choose some model. Let's make a set of predictions for that one validation point by using say 100 different random samples for training the model. So we are going to get 100 y values. The difference between the mean of those values and the true value is called the bias. The variance of the distribution is the variance.

Depending on what model we use we can trade off between these two. Let's consider the two extremes. The lowest variance model is one where completely ignore the data. Let's say we simply predict 42 for every x. That model has zero variance across different training samples at every point. However it is clearly biased. The bias is simply 42-y_v.

One the other extreme we can choose a model which overfits as much as possible. For example fit a 100 degree polynomial to 100 data points. Or alternatively, linearly interpolate between nearest neighbors. This has low bias. Why? Because for any random sample the neighboring points to x_v will fluctuate widely but they will interpolate higher just about as often as they will interpolate low. So on average across the samples, they will cancel out and the bias will therefore be very low unless the true curve has lots of high frequency variation.

Hoever these overfit models have large variance across the random samples because they are not smoothing the data. The interpolation model just uses two data points to predict the intermediate one and these therefore create a lot of noise.

Note that the bias is measured at a single point. It doesn't matter if it is positive or negative. It is still a bias at any given x. The biases averaged over all the x values will probably be small but that doesn't make it unbiased.

One more example. Say you are trying to predict the temperature at set of locations in the US at some time. Let's assume you have 10,000 training points. Again, you can get a low variance model by doing something simple by just returning the average. But this will be biased low in the state of Florida and biased high in the state of Alaska. You'd be better if you used the average for each state. But even then, you will be biased high in the winter and low in the summer. So now you include the month in your model. But you're still going to be biased low in Death Valley and high on Mt Shasta. So now you go to the zip code level of granularity. But eventually if you keep doing this to reduce the bias, you run out of data points. Maybe for a given zip code and month, you have only one data point. Clearly this is going to create lots of variance. So you see having a more complicated model lowers the bias at the expense of variance.

So you see there is a trade off. Models that are smoother have lower variance across training samples but don't capture the real shape of the curve as well. Models that are less smooth can better capture the curve but at the expense of being noisier. Somewhere in the middle is a Goldilocks model that makes an acceptable tradeoff between the two.


Imagine if model building task could be repeated for different training datasets, i.e. we train a new model for different dataset every time(shown in the figure below). If we fix a test data point and evaluate model prediction on this point, the predictions will be varied due to randomness in the model generation process. From the below figure for this situation, P_1, P_2, …, P_n are different predictions and random too. enter image description here

Let the mean of predictions be -

enter image description here

Bias Error is due to the difference between mean of these predictions and the correct value. enter image description here

Variance Error is nothing but the variance in these predictions, i.e. how varied are these predictions. enter image description here

This is the intuition behind bias and variance error.

For detailed explanation visit right intuition behind bias variance tradeoff


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