- A pivot is a function of the sample data and unknown parameter
- Pivot estimation can be used to construct intervals known as confidence intervals
- Something important to remember is that its distribution does not depend on an unknown parameter
2025-06-08
Let \(X_1, X_2, \dots, X_n\) be a sample with parameter \(\theta\)
A pivot is a statistic \(T(X_1, \dots, X_n; \theta)\) such that:
\[ \text{Distribution of } T \text{ does not depend on } \theta \]
This allows us to derive probabilistic statements about \(\theta\)
Let’s say \(X_1, X_2, \dots, X_n\) be a sample with the parameter \(\theta\)
A pivot is a statistic \(T(X_1, \dots, X_n; \theta)\) such that:
\[ \text{Distribution of } T \text{ does not depend on } \theta \]
This allows us to derive probabilistic statements about \(\theta\)
Let’s suppose \(X_1, \dots, X_n \sim \mathcal{N}(\mu, \sigma^2)\) with a known \(\sigma\)
Pivot: \[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim \mathcal{N}(0, 1) \]
Then it can be rearranged to form a confidence interval: \[ \mu \in \left[ \bar{X} - z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}, \, \bar{X} + z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} \right] \]
## [1] -0.2579979
So what do all these values mean? Let’s take a look…
Here, we have simulated a sample of size n = 30 from a normal distribution with mean of μ = 5 and a standard deviation of σ = 2
The Z-statistic is a pivot: \[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \]
Thus, from this we can tell exactly how many standard errors from the sample mean \(\bar{X}\) is from the true mean \(\mu\)
This visualization also accomplishes showing how the Z-score changes as both the sample size and mean change
For LARGER values of n it would decrease the value of standard error, thus making the Z-score increase even with the smallest changes in deviation from the population mean
Pivot methods are a bridge between theorotical and practical estimations
Thank you for watching! - Andrew Olariu