# Linear regression with non-symmetric cost function?

I want to predict some value $Y(x)$ and I am trying to get some prediction $\hat Y(x)$ that optimizes between being as low as possible, but still being larger than $Y(x)$. In other words: $$\text{cost}\left\{ Y(x) \gtrsim \hat Y(x) \right\} >> \text{cost}\left\{ \hat Y(x) \gtrsim Y(x) \right\}$$

I think a simple linear regression should do totally fine. So I somewhat know how to implement this manually, but I guess I'm not the first one with this kind of problem. Are there any packages/libraries (preferably python) out there doing what I want to do? What's the keyword I need to look for?

What if I knew a function $Y_0(x) > 0$ where $Y(x) > Y_0(x)$. What's the best way to implement these restrictions?

• Probably, the most simple solution is to use different weights, based on whether the prediction is positive or negative. I should have thought of that earlier. – asPlankBridge Mar 2 '16 at 2:23

If I understand you correctly, you want to err on the side of overestimating. If so, you need an appropriate, asymmetric cost function. One simple candidate is to tweak the squared loss:

$\mathcal L: (x,\alpha) \to x^2 \left( \mathrm{sgn} x + \alpha \right)^2$

where $-1 < \alpha < 1$ is a parameter you can use to trade off the penalty of underestimation against overestimation. Positive values of $\alpha$ penalize overestimation, so you will want to set $\alpha$ negative. In python this looks like def loss(x, a): return x**2 * (numpy.sign(x) + a)**2 Next let's generate some data:

import numpy
x = numpy.arange(-10, 10, 0.1)
y = -0.1*x**2 + x + numpy.sin(x) + 0.1*numpy.random.randn(len(x)) Finally, we will do our regression in tensorflow, a machine learning library from Google that supports automated differentiation (making gradient-based optimization of such problems simpler). I will use this example as a starting point.

import tensorflow as tf

X = tf.placeholder("float") # create symbolic variables
Y = tf.placeholder("float")

w = tf.Variable(0.0, name="coeff")
b = tf.Variable(0.0, name="offset")
y_model = tf.mul(X, w) + b

cost = tf.pow(y_model-Y, 2) # use sqr error for cost function
def acost(a): return tf.pow(y_model-Y, 2) * tf.pow(tf.sign(y_model-Y) + a, 2)

sess = tf.Session()
init = tf.initialize_all_variables()
sess.run(init)

for i in range(100):
for (xi, yi) in zip(x, y):
#         sess.run(train_op, feed_dict={X: xi, Y: yi})
sess.run(train_op2, feed_dict={X: xi, Y: yi})

print(sess.run(w), sess.run(b))


cost is the regular squared error, while acost is the aforementioned asymmetric loss function.

If you use cost you get

1.00764 -3.32445 If you use acost you get

1.02604 -1.07742 acost clearly tries not to underestimate. I did not check for convergence, but you get the idea.

• Thank you for this detailed answer: One question to the definition of the acost function though. Does it matter that you calculate y_model-Y twice? – asPlankBridge Mar 2 '16 at 6:15
• You mean in terms of speed? I don't know; you'll have to time it yourself to see if tensorflow avoids recalculation. It is fine otherwise. – Emre Mar 2 '16 at 6:26

Pick an asymmetric loss function. One option is quantile regression (linear but with different slopes for positive and negative errors).