5.2 Principles of Least Squares, Most Probable Value, and Weighted Observations

Principles of Least Squares, Most Probable Value, and Weighted Observations

When surveying measurements contain redundant observations (more measurements than the minimum necessary to determine unknown geometric parameters), small inconsistencies known as misclosures occur due to random errors. The Principle of Least Squares provides the mathematically rigorous framework for resolving these discrepancies to obtain the Most Probable Value (MPV) of unknown quantities and establish unbiased estimates of measurement precision.

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1. Fundamental Principle of Least Squares

1.1 The Least-Squares Postulate

The mathematical principle of least squares, independently developed by Carl Friedrich Gauss and Adrien-Marie Legendre, states that for a system of redundant independent observations of equal precision, the most probable values of the unknown parameters are those that minimize the sum of the squares of the observation residuals:

$\sum_{i=1}^n v_i^2 = v_1^2 + v_2^2 + \dots + v_n^2 = \text{minimum}$

where $v_i = \text{Adjusted Value} - \text{Observed Value} = f(\hat{X}) - l_i$.

1.2 Mathematical Proof: Arithmetic Mean as MPV

Consider $n$ independent observations $l_1, l_2, \dots, l_n$ of a single unknown quantity $x$ with equal weight. The residuals are $v_i = x - l_i$. The sum of squared residuals function $S(x)$ is: $S(x) = \sum_{i=1}^n v_i^2 = \sum_{i=1}^n (x - l_i)^2$

To find the value of $x$ that minimizes $S(x)$, take the first derivative with respect to $x$ and equate to zero: $\frac{dS}{dx} = \frac{d}{dx} \left[ \sum_{i=1}^n (x - l_i)^2 \right] = 2 \sum_{i=1}^n (x - l_i) = 0$ $\sum_{i=1}^n x - \sum_{i=1}^n l_i = 0 \implies n x = \sum_{i=1}^n l_i \implies x = \frac{\sum_{i=1}^n l_i}{n} = \bar{x}$

Thus, the arithmetic mean is mathematically proven to be the least-squares most probable value for direct observations of equal weight.

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2. Weights of Observations

In real-world geodetic operations, observations rarely possess identical precision. Measurements performed with different instruments, over varying line lengths, under diverse weather conditions, or by observers of unequal skill must be weighted according to their relative reliability.

2.1 Definition and Variance Relationship

The weight ($w_i$ or $p_i$) of an observation is a relative dimensionless number inversely proportional to its variance ($\sigma_i^2$):

$w_i \propto \frac{1}{\sigma_i^2} \implies w_i = \frac{\sigma_0^2}{\sigma_i^2}$

where $\sigma_0^2$ is the reference variance (or variance of an observation of unit weight, $w=1$).

From this definition, the fundamental weight-variance relation is established: $w_1 \sigma_1^2 = w_2 \sigma_2^2 = \dots = w_n \sigma_n^2 = \sigma_0^2$

Key implications:

• High variance $\implies$ low precision $\implies$ low weight.
• An observation with twice the standard deviation has one-fourth the weight ($w \propto 1/s^2$).

2.2 Weight Determination Rules in Surveying Practice

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3. Weighted Mean and Precision Metrics

For $n$ observations of unequal weight $l_1, l_2, \dots, l_n$ with assigned weights $w_1, w_2, \dots, w_n$, the least-squares objective function becomes:

$\sum_{i=1}^n w_i v_i^2 = w_1 v_1^2 + w_2 v_2^2 + \dots + w_n v_n^2 = \text{minimum}$

3.1 Weighted Mean ($\bar{x}_w$) Formula

Differentiating $\sum w_i (x - l_i)^2 = 0$ yields the weighted mean: $\bar{x}_w = \frac{\sum_{i=1}^n w_i l_i}{\sum_{i=1}^n w_i} = \frac{w_1 l_1 + w_2 l_2 + \dots + w_n l_n}{w_1 + w_2 + \dots + w_n}$

3.2 Standard Error of Unit Weight ($\sigma_0$)

The standard error of unit weight evaluates overall model consistency and residual dispersion: $\sigma_0 = \sqrt{\frac{\sum_{i=1}^n w_i v_i^2}{n - u}} = \sqrt{\frac{\sum_{i=1}^n w_i v_i^2}{r}}$ where $n$ is the number of observations, $u$ is the number of unknown parameters, and $r = n - u$ is the degrees of freedom (redundancy).

3.3 Standard Error of an Individual Observation of Weight $w_i$

$s_{x_i} = \frac{\sigma_0}{\sqrt{w_i}}$

3.4 Standard Error of the Weighted Mean ($s_{\bar{x}_w}$)

$s_{\bar{x}_w} = \frac{\sigma_0}{\sqrt{\sum_{i=1}^n w_i}} = \sqrt{\frac{\sum w_i v_i^2}{(n - 1) \sum w_i}}$

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Worked Numerical Example 1: Weighted Adjustment of Differential Level Lines

Three differential level lines are run from established benchmark BM-A (Elevation = $125.400\text{ m}$) to determine the elevation of point B. The observed elevation differences, route distances, and calculated weights are summarized below:

• Line 1: $\Delta h_1 = +14.250\text{ m}$, Distance $D_1 = 1.0\text{ km} \implies \text{Observed Elevation } E_1 = 139.650\text{ m}$
• Line 2: $\Delta h_2 = +14.280\text{ m}$, Distance $D_2 = 2.0\text{ km} \implies \text{Observed Elevation } E_2 = 139.680\text{ m}$
• Line 3: $\Delta h_3 = +14.230\text{ m}$, Distance $D_3 = 4.0\text{ km} \implies \text{Observed Elevation } E_3 = 139.630\text{ m}$

Step 1: Assign Relative Weights Using $w_i = \frac{1}{D_i}$:

• $w_1 = \frac{1}{1.0} = 1.000$
• $w_2 = \frac{1}{2.0} = 0.500$
• $w_3 = \frac{1}{4.0} = 0.250$ $\sum w_i = 1.000 + 0.500 + 0.250 = 1.750$

Step 2: Compute Weighted Mean Elevation ($\bar{E}_B$) $\sum w_i E_i = (1.000 \cdot 139.650) + (0.500 \cdot 139.680) + (0.250 \cdot 139.630) = 139.650 + 69.840 + 34.9075 = 244.3975$ $\bar{E}_B = \frac{\sum w_i E_i}{\sum w_i} = \frac{244.3975}{1.750} = 139.6557\text{ m}$

Step 3: Calculate Residuals ($v_i = \bar{E}_B - E_i$) and Weighted Squares ($w_i v_i^2$)

• $v_1 = 139.6557 - 139.6500 = +0.0057\text{ m} \implies w_1 v_1^2 = 1.000 \cdot (0.0057)^2 = 0.00003249$
• $v_2 = 139.6557 - 139.6800 = -0.0243\text{ m} \implies w_2 v_2^2 = 0.500 \cdot (-0.0243)^2 = 0.000295245$
• $v_3 = 139.6557 - 139.6300 = +0.0257\text{ m} \implies w_3 v_3^2 = 0.250 \cdot (+0.0257)^2 = 0.000165123$ $\sum w_i v_i^2 = 0.00003249 + 0.000295245 + 0.000165123 = 0.000492858\text{ m}^2$

Step 4: Compute Standard Error of Unit Weight ($\sigma_0$) Here $n=3$ observations, $u=1$ unknown ($\bar{E}_B$), redundancy $r = 3 - 1 = 2$: $\sigma_0 = \sqrt{\frac{0.000492858}{2}} = \sqrt{0.000246429} = \pm 0.01570\text{ m} = \pm 15.7\text{ mm}$

Step 5: Compute Standard Error of Adjusted Elevation ($\sigma_{\bar{E}_B}$) $\sigma_{\bar{E}_B} = \frac{\sigma_0}{\sqrt{\sum w_i}} = \frac{0.01570}{\sqrt{1.750}} = \frac{0.01570}{1.322876} = \pm 0.01187\text{ m} \approx \pm 11.9\text{ mm}$

The adjusted elevation of point B is $139.6557\text{ m} \pm 0.0119\text{ m}$.

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4. Normal Equations for Multiple Unknown Parameters

When observations are indirect functions of multiple unknown parameters $X = [x, y, z, \dots]^T$, observation equations are formed:

$v_i = a_i x + b_i y + c_i z + \dots - l_i$

where $a_i, b_i, c_i$ are coefficient partial derivatives and $l_i = l_{observed, i} - l_{computed, i}^0$.

4.1 Algebraic Normal Equations Structure

To minimize $\sum w_i v_i^2$, partial derivatives of $\sum w_i (a_i x + b_i y + c_i z - l_i)^2$ with respect to each unknown $x, y, z$ are set to zero, yielding the system of Normal Equations:

$$\begin{aligned} ([a a w]) x + ([a b w]) y + ([a c w]) z &= [a l w] \\ ([a b w]) x + ([b b w]) y + ([b c w]) z &= [b l w] \\ ([a c w]) x + ([b c w]) y + ([c c w]) z &= [c l w] \end{aligned}$$

where Gaussian summation bracket notation defines matrix coefficients: $[a a w] = \sum_{i=1}^n a_i^2 w_i, \quad [a b w] = \sum_{i=1}^n a_i b_i w_i, \quad [a l w] = \sum_{i=1}^n a_i l_i w_i$.

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Worked Numerical Example 2: Normal Equations Solution for Two Unknown Station Heights

Determine the elevations of two benchmark stations $h_A$ and $h_B$ relative to control BM-0 ($Elev = 0.000\text{ m}$) given four observed elevation differences of equal weight ($w_i = 1.0$):

1. $l_1 = +10.05\text{ m}$ (from BM-0 to A) $\implies h_A = 10.05$
2. $l_2 = +15.10\text{ m}$ (from BM-0 to B) $\implies h_B = 15.10$
3. $l_3 = +5.00\text{ m}$ (from A to B) $\implies h_B - h_A = 5.00$
4. $l_4 = +15.02\text{ m}$ (from BM-0 to B) $\implies h_B = 15.02$

Step 1: Set Up Observation Equations ($v_i = a_i h_A + b_i h_B - l_i$)

• Line 1: $v_1 = (1) h_A + (0) h_B - 10.05$
• Line 2: $v_2 = (0) h_A + (1) h_B - 15.10$
• Line 3: $v_3 = (-1) h_A + (1) h_B - 5.00$
• Line 4: $v_4 = (0) h_A + (1) h_B - 15.02$

Step 2: Calculate Summation Terms

• $[a a] = (1)^2 + (0)^2 + (-1)^2 + (0)^2 = 1 + 0 + 1 + 0 = 2$
• $[b b] = (0)^2 + (1)^2 + (1)^2 + (1)^2 = 0 + 1 + 1 + 1 = 3$
• $[a b] = (1)(0) + (0)(1) + (-1)(1) + (0)(1) = 0 + 0 - 1 + 0 = -1$
• $[a l] = (1)(10.05) + (0)(15.10) + (-1)(5.00) + (0)(15.02) = 10.05 - 5.00 = 5.05$
• $[b l] = (0)(10.05) + (1)(15.10) + (1)(5.00) + (1)(15.02) = 15.10 + 5.00 + 15.02 = 35.12$

Step 3: Construct System of Normal Equations

$$\begin{cases} 2 h_A - 1 h_B = 5.05 \\ -1 h_A + 3 h_B = 35.12 \end{cases}$$

Step 4: Solve System of Equations From Equation 1: $2 h_A = 5.05 + h_B \implies h_A = 2.525 + 0.5 h_B$. Substitute into Equation 2: $- (2.525 + 0.5 h_B) + 3 h_B = 35.12$ $-2.525 + 2.5 h_B = 35.12 \implies 2.5 h_B = 37.645 \implies h_B = 15.058\text{ m}$ Substituting back for $h_A$: $h_A = 2.525 + 0.5(15.058) = 2.525 + 7.529 = 10.054\text{ m}$

Final Most Probable Values: $h_A = 10.054\text{ m}, \quad h_B = 15.058\text{ m}$