Suppose I have a Gaussian random variable $X \sim N(\mu,\lambda^{-1})$ with $\lambda > 1$. I know $\lambda$ but the mean is unknown. I draw once from this tribulation and use the observation with, for example, Bayes' rule and a prior to form a belief of the mean of the distribution. Alternatively, suppose the random variable was instead distributed according to $N(\mu,1)$, ie. with a higher variance, but I was able to draw $n > 1$ times from it. Is there a relationship between $\lambda$ and $n$ that I can exploit to determine when these two scenarios would be equivalent?

Intuitively, it seems to me that drawing once from a tighter distribution should be equivalent to drawing multiple times from a wider distribution, but I can't quite put my finger on how I'd go about trying to derive a relationship.

  • $\begingroup$ 1. In English, "tribulation" doesn't have a relevant meaning for this context. Do you mean "distribution"? 2. Have you considered comparing the sampling distrfibution of sample means for iid samples from each of those two normal distributions? $\endgroup$ – Glen_b -Reinstate Monica Jul 18 '18 at 3:56

They are equivalent if you are only interested in inference for $\mu$.

Since the variance is known, a $95\%$ confidence interval for the mean (assuming observations are independent) is given by

$$\bar X \pm \frac{z_{\alpha/2}}{\sqrt{\lambda n}}$$

As long as $\lambda_1 n_1 = \lambda_2 n_2$, then in this simple example we can argue that the samples contain the same amount of information for $\mu$.

To directly answer your question, set $n_1 = 1$ and $\lambda_1 = \lambda$. Then we can write $$\lambda_2 = \lambda \frac{1}{n_2}$$ So with a second data point, you would need to halve $\lambda$, etc. In a Bayesian setting, the relationship may change according to your prior, but a relationship will exist nonetheless.

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