Konfidenzintervall für das Produkt zweier Parameter

Nehmen wir an, wir haben zwei Parameter, $p_1$ und $p_2$ . Wir haben auch zwei Maximum-Likelihood-Schätzer $\hat{p_1}$ und $\hat{p_2}$ und zwei Konfidenzintervalle für diese Parameter. Gibt es eine Möglichkeit, ein Konfidenzintervall für zu erstellen $p_1p_2$?

Mit der Delta-Methode können Sie den Standardfehler von $\hat{p_{1}}\hat{p_{2}}$ berechnen . Das Delta-Verfahren besagt, dass eine Annäherung der Varianz einer Funktion $g(t)$ gegeben ist durch:

$$\mathrm{Var}(g(t))\approx \sum_{i=1}^{k}g'_{i}(\theta)^{2}\mathrm{Var}(t_{i})+2\sum_{i>j}g'_{i}(\theta)g'_{j}(\theta)\mathrm{Cov}(t_{i},t_{j})$$ , p Die Annäherung der Erwartung von$g(t)$ist andererseits gegeben durch: $$\mathrm{\mathbf{E}}(g(t))\approx g(\theta)$$ Die Erwartung ist also einfach die Funktion. Ihre Funktion$g(t)$ist:$g(p_{1}, p_{2})=p_{1}p_{2}$ . Die Erwartung von $g(p_{1}, p_{2})=p_{1}p_{2}$ wäre einfach:$p_{1}p_{2}$ . Für die Varianz benötigen wir die partiellen Ableitungen von$g(p_{1}, p_{2})$ : $$\begin{align} \frac{\partial}{\partial p_{1}}g(p_{1}p_{2}) & = p_{2} \\ \frac{\partial}{\partial p_{2}}g(p_{1}p_{2}) &= p_{1} \\ \end{align}$$

Using the function for the variance above, we get:

$$\mathrm{Var}(\hat{p_{1}}\hat{p_{2}})=\hat{p_{2}}^{2}\mathrm{Var}(\hat{p_{1}}) + \hat{p_{1}}^{2}\mathrm{Var}(\hat{p_{2}})+2\cdot \hat{p_{1}}\hat{p_{2}}\mathrm{Cov}(\hat{p_{1}},\hat{p_{2}})$$ The standard error would then simply be the sqare root of the above expression. Once you've got the standard error, it is straight forward to calculate a 95% confidence interval for $\hat{p_{1}}\hat{p_{2}}$: $\hat{p_{1}}\hat{p_{2}}\pm 1.96\cdot \widehat{\mathrm{SE}}(\hat{p_{1}}\hat{p_{2}})$

To caluclate the standard error of $\hat{p_{1}}\hat{p_{2}}$, you need the variance of $\hat{p_{1}}$ and $\hat{p_{2}}$ which you usually can get by the variance-covariance matrix $\Sigma$ which would be a 2x2-matrix in your case because you have two estimates. The diagonal elements in the variance-covariance matrix are the variances of $\hat{p_{1}}$ and $\hat{p_{2}}$ while the off-diagonal elements are the covariance of $\hat{p_{1}}$ and $\hat{p_{2}}$ (the matrix is symmetric). As @gung mentions in the comments, the variance-covariance matrix can be extracted by most statistical softwares. Sometimes, estimation algorithms provide the Hessian matrix (I won't go into details about that here), and the variance-covariance matrix can be estimated by the inverse of the negative Hessian (but only if you maximized the log-likelihood!; see this post). Again, consult the documentation of your statistical software and/or the web on how to extract the Hessian and on how to calculate the inverse of a matrix.

Alternatively, you can get the variances of $\hat{p_{1}}$ and $\hat{p_{2}}$ from the confidence intervals in the following way (this is valid for a 95%-CI): $\mathrm{SE}(\hat{p_{1}})=(\text{upper limit} - \text{lower limit})/3.92$. For an $100(1-\alpha)\%$-CI, the estimated standard error is: $\mathrm{SE}(\hat{p_{1}})=(\text{upper limit} - \text{lower limit})/(2\cdot z_{1-\alpha/2})$, where $z_{1-\alpha/2}$ is the $(1-\alpha/2)$ quantile of the standard normal distribution (for $\alpha=0.05$, $z_{0.975}\approx 1.96$). Then, $\mathrm{Var}(\hat{p_{1}}) = \mathrm{SE}(\hat{p_{1}})^{2}$. The same is true for the variance of $\hat{p_{2}}$. We need to covariance of $\hat{p_{1}}$ and $\hat{p_{2}}$ too (see paragraph above). If $\hat{p_{1}}$ and $\hat{p_{2}}$ are independent, the covariance is zero and we can drop the term.

This paper might provide additional information.

I found a different equation for calculation of variance of product.

If x and y are independently distributed, the variance of the product is relatively straightforward: V(x*y)= V(y)*E(x)^2 + V(x)*E(y)^2 + V(x)*V(y) These results also generalize to cases involving three or more variables (Goodman 1960). Source: Regulating Pesticides (1980), appendix F

Coolserdash: The last component V(x)*V(y) is missing in your equation. Is the referenced book (Regulating Pesticides) wrong?

Also, both equations might not be perfect. "... we show that the distribution of the product of three independent normal variables is not normal." (source). I would expect some positive skew even in the product of two normally distributed variables.

1. The length of the CI / 2 / 1.96 = se, i.e. the standard error of A or B
2. se^2 = var, i.e. the variance of the estimate A or B
3. Use the estimated A or B as the means of A or B, i.e. E(A) or E(B)
4. Follow this page http://falkenblog.blogspot.se/2008/07/formula-for-varxy.html to get var(A*B), i.e. var(C)
5. Square-root of var(C) is the se of C
6. (C - 1.96*se(C), C + 1.96*se(C)) is the 95% CI of C

Note that if your A and B are correlated, you need to consider their covariance as well.