Why Measure First-Stage Strength?
In the scalar instrumental-variables model, the instruments identify the endogenous regressor through the part of that regressor predicted by the first stage. The first-stage coefficient vector alone does not tell the whole story: its scale depends on how the instruments are measured, and the same coefficient vector can provide more information in a larger sample. The concentration parameter combines instrument strength, the design, and first-stage noise in a single scale-free quantity.
The parameter is useful because it describes the strength of the projected first-stage signal relative to its noise. It therefore indexes how informative the instruments are for identifying the structural coefficient. It is not, by itself, a standard error, a test statistic, or a guarantee that any particular approximation to IV inference is accurate.
Definition in the Homoskedastic First-Stage Model
Consider \(n\) observations and \(q\) excluded instruments. Write the first stage as \(x=Z\pi+v\), where \(x\) is the \(n\)-vector of observations on the scalar endogenous regressor, \(Z\) is a fixed \(n\)-by-\(q\) instrument matrix of full column rank, and \(\pi\) is the \(q\)-vector of first-stage coefficients. The first-stage disturbance satisfies \(E[v\mid Z]=0\) and \(\operatorname{Var}(v\mid Z)=\sigma_v^2 I_n\), with \(0<\sigma_v^2<\infty\). If controls or an intercept are included, the same definition applies after the relevant variables have been residualized with respect to those controls.
The numerator is the squared length of the systematic first-stage component \(Z\pi\); the denominator is the variance of the first-stage disturbance for one observation. Thus \(\mu^2\) compares the total fitted signal in the sample with the disturbance variance. The definition does not require normal errors. Normality will be imposed below only when deriving an exact sampling distribution.
Worked Example: Computing Concentration for a Fixed Design
Suppose there are \(n=100\) observations and two instruments, with \(Z'Z=100I_2\), \(\pi=(0.2,0.1)'\), and \(\sigma_v^2=0.25\). Compute the concentration parameter.
First calculate the quadratic form: $$ \pi'Z'Z\pi =100\pi'\pi =100\bigl(0.2^2+0.1^2\bigr) =100(0.05) =5. $$ Then divide by the first-stage disturbance variance: $$ \mu^2=\frac{5}{0.25}=20. $$ So the concentration parameter is \(20\). This number is not a percentage and is not itself a first-stage test statistic; it records the scale of the projected signal relative to disturbance variance.
Projection Geometry and an Exact Distribution
Let \(P_Z=Z(Z'Z)^{-1}Z'\), the projection matrix onto the column space of the instruments. The first-stage fitted values are \(P_Zx\). Since \(P_ZZ\pi=Z\pi\), their squared length is $$ x'P_Zx=(Z\pi+v)'P_Z(Z\pi+v). $$ The expectation of this quantity is the sum of systematic signal and projected disturbance energy.
Proof. Set \(U=Z(Z'Z)^{-1/2}\), so \(U'U=I_q\) and \(P_Z=UU'\). It follows that $$ \frac{x'P_Zx}{\sigma_v^2} =\left\|\frac{U'x}{\sigma_v}\right\|^2. $$ Because \(x=Z\pi+v\), $$ \frac{U'x}{\sigma_v} =\frac{(Z'Z)^{1/2}\pi}{\sigma_v}+\frac{U'v}{\sigma_v}. $$ Under the stated normality assumption, \(U'v/\sigma_v\sim N(0,I_q)\). The vector on the right is therefore a \(q\)-dimensional standard normal vector shifted by a fixed vector whose squared length is $$ \left\|\frac{(Z'Z)^{1/2}\pi}{\sigma_v}\right\|^2 =\frac{\pi'Z'Z\pi}{\sigma_v^2} =\mu^2. $$ By the definition of the noncentral chi-square distribution, its squared length has distribution \(\chi^2_q(\mu^2)\). To obtain the moments, write that squared length as \(\sum_{j=1}^q(G_j+a_j)^2\), where the \(G_j\) are independent standard normal variables and \(\sum_j a_j^2=\mu^2\). Each term has expectation \(1+a_j^2\) and variance \(2+4a_j^2\). Summing gives expectation \(q+\mu^2\) and variance \(2q+4\mu^2=2(q+2\mu^2)\), as claimed. \(\square\)
This result makes the interpretation concrete. The projection includes \(q\) directions of first-stage noise, contributing \(q\) to the expected normalized projected energy. The additional contribution \(\mu^2\) is the squared signal in those same directions, measured in units of disturbance variance.
Worked Example: Projected First-Stage Energy
Continue with \(q=2\) and \(\mu^2=20\), and suppose the first-stage errors are normal as in the theorem. The normalized projected energy has distribution \(\chi^2_2(20)\). Its expectation is $$ E\left[\frac{x'P_Zx}{\sigma_v^2}\middle|Z\right] =2+20=22, $$ and its variance is $$ \operatorname{Var}\left(\frac{x'P_Zx}{\sigma_v^2}\middle|Z\right) =2(2+2\cdot20)=84. $$ The expectation is not just the concentration parameter: it also includes the two degrees of freedom of projected noise. This distinction matters when interpreting projected first-stage quantities.
How Sample Size and Instrument Strength Enter
For a fixed design, the definition already gives the sample-size relationship: the value depends on the total signal \(\pi'Z'Z\pi\), not merely on \(\pi\). To make the role of sample size explicit, suppose rows of \(Z\) are independently sampled, centered instrument vectors with covariance matrix \(\Sigma_Z\), and \(\pi\) and \(\sigma_v^2\) are fixed. The conditional concentration parameter is still computed from the realized \(Z\). Averaging over random designs gives $$ E_Z[\mu^2] =\frac{\pi'E[Z'Z]\pi}{\sigma_v^2} =\frac{n\,\pi'\Sigma_Z\pi}{\sigma_v^2}, $$ since \(E[Z'Z]=n\Sigma_Z\). Thus, on average, concentration grows linearly with \(n\) when the population first-stage signal and disturbance variance are held fixed.
The term \(\pi'\Sigma_Z\pi\) is the variance of the linear instrument signal for one observation. A larger first-stage signal raises concentration; a larger first-stage disturbance variance lowers it. The instrument scale is accounted for through \(\Sigma_Z\), which is important because coefficient size by itself is sensitive to units.
Worked Example: Comparing Two Sample Sizes
Suppose the instruments have covariance matrix \(\Sigma_Z=\operatorname{diag}(4,1)\), the first-stage coefficients are \(\pi=(0.1,0.2)'\), and \(\sigma_v^2=0.5\). Find the expected concentration for \(n=50\) and \(n=200\).
The per-observation signal is $$ \pi'\Sigma_Z\pi =4(0.1)^2+1(0.2)^2 =0.04+0.04 =0.08. $$ For \(n=50\), $$ E_Z[\mu^2]=\frac{50(0.08)}{0.5}=8. $$ For \(n=200\), $$ E_Z[\mu^2]=\frac{200(0.08)}{0.5}=32. $$ The sample size is four times as large in the second case, and the expected concentration is also four times as large. These are expectations over random designs; a particular realized instrument matrix can yield a different value.
Why It Indexes IV Precision
In the scalar structural equation \(y=\beta x+u\), the IV estimator uses instrument-predicted variation in \(x\) to distinguish the structural coefficient \(\beta\) from the effect of the endogenous disturbance \(u\). The concentration parameter describes the size of that predicted variation relative to first-stage noise. When the projected signal is larger relative to noise, the instruments generally provide more information about the structural coefficient; when it is small, the identifying variation is difficult to distinguish from first-stage noise.
The connection can also be seen in the denominator of the just-identified or overidentified 2SLS expression: its first-stage component is governed by \(x'P_Zx\). The theorem shows that this projected quantity has a noncentrality parameter equal to \(\mu^2\). However, \(\mu^2\) does not determine the full sampling distribution of the IV estimator by itself. That distribution also depends on the structural disturbance and its relationship with the first-stage disturbance. The concentration parameter is an index of first-stage identifying information, not a complete description of inferential uncertainty.
Worked Example: Instrument Rescaling Does Not Change Concentration
Suppose \(Z'Z=100I_2\), \(\pi=(0.2,0.1)'\), and \(\sigma_v^2=0.25\), so the concentration is \(20\). Now rescale the first instrument by two, leaving the second unchanged. Write \(Z^*=ZA\), where \(A=\operatorname{diag}(2,1)\). To leave the first-stage signal unchanged, the new coefficient vector must be \(\pi^*=A^{-1}\pi=(0.1,0.1)'\).
The transformed Gram matrix is $$ Z^{*'}Z^*=A'Z'ZA =\operatorname{diag}(400,100). $$ Thus the transformed signal quadratic form is $$ \pi^{*'}Z^{*'}Z^*\pi^* =400(0.1)^2+100(0.1)^2 =4+1=5. $$ The original quadratic form was also $$ \pi'Z'Z\pi=100(0.2)^2+100(0.1)^2=4+1=5. $$ Dividing by \(0.25\) gives concentration \(20\) in either parameterization. The numerical first-stage coefficient changed, but the fitted signal and its strength did not.
Common Mistake and Key Takeaway
Mistake: “The first-stage coefficient is small, so the instruments must be weak,” or conversely, “the coefficient is large, so the instruments must be strong.” This reasoning ignores the units and variability of the instruments, the sample design, and the disturbance variance. Rescaling an instrument changes its coefficient while leaving the fitted signal unchanged.
Correction: Evaluate the signal through \(\pi'Z'Z\pi\) and compare it with \(\sigma_v^2\). The concentration parameter incorporates the design and noise as well as the first-stage coefficient. It helps index the precision of IV identification, but should not be mistaken for a universal cutoff or a complete measure of IV uncertainty.
Check Your Understanding
Use the definition and assumptions in this tutorial to answer the following.
- In the fixed-design model, what features determine whether the concentration parameter is zero?
- Why does the exact noncentral chi-square result require normal first-stage errors, while the definition of concentration does not?
- If the first-stage signal per observation and disturbance variance are fixed, how does expected concentration change when sample size doubles?
- Why can rescaling an instrument change its first-stage coefficient without changing concentration?
- Does knowing \(\mu^2\) alone determine the full sampling uncertainty of the IV estimator? Explain.