# Kyle's Lambda

| Last updated on | 2 min read

A measure of market impact cost from Kyle (1985), which can be interpreted as the cost of demanding a certain amount of liquidity over a given time period.

## Definition

Following Hasbrouck (2009) and Goyenko, Holden, Trzcinka (2009), Kyle's Lambda for a given stock $i$ and day $t$, is calculated as the slope coefficient $\lambda_{i,t}$ in the regression:

$ret_{i,t,n}= \delta_{i,t} + \lambda_{i,t} S_{i,t,n}+\epsilon_{i,t,n}$

where for the $n$th five-minute period on date $t$ and stock $i$, $ret_{i,t,n}$ is the stock return and $S_{i,t,n}$ is the sum of the signed square-root dollar volume, that is,

$S_{i,t,n}=\sum_k{\text{sign}}(dvol_{i,t,n,k}) \sqrt{dvol_{i,t,n,k}}$

## Source Code This example Python code is not optimized for speed and serves only demonstration purpose. It may contain errors.

It returns $\lambda \times 10^6$

# KylesLambda.py
import numpy as np

name = 'KylesLambda'
description = """
A measure of market impact cost from Kyle (1985),
which can be interpreted as the cost of demanding a certain amount of liquidity over a given time period.
Result is Lambda*1E6.
"""
vars_needed = ['Price', 'Volume', 'Direction']

def estimate(data):
price = data['Price'].to_numpy()
volume = data['Volume'].to_numpy()
direction = data['Direction'].to_numpy()
sqrt_dollar_volume = np.sqrt(np.multiply(price, volume))
signed_sqrt_dollar_volume = np.abs(
np.multiply(direction, sqrt_dollar_volume))
# Find the total signed sqrt dollar volume and return per 5 min.
timestamps = np.array(data.index, dtype='datetime64')
last_ts, last_price = timestamps, price
bracket_ssdv = 0
bracket = last_ts + np.timedelta64(5, 'm')
rets, ssdvs, = [], []
for idx, ts in enumerate(timestamps):
if ts <= bracket:
bracket_ssdv += signed_sqrt_dollar_volume[idx]
else:
ret = np.log(price[idx-1]/last_price)
if not np.isnan(ret) and not np.isnan(bracket_ssdv):
rets.append(ret)
ssdvs.append(bracket_ssdv)
# Reset bracket
bracket = ts + np.timedelta64(5, 'm')
last_price = price[idx]
bracket_ssdv = signed_sqrt_dollar_volume[idx]
# Perform regression.
x = np.vstack([np.ones(len(ssdvs)), np.array(ssdvs)]).T
try:
coef, _, _, _ = np.linalg.lstsq(x, np.array(rets), rcond=None)
except np.linalg.LinAlgError:
return None
else:
return None if np.isnan(coef) else coef*1E6