1/E(X) is not E(1/X)

When modeling with uncertainty we often care about the expected value of our result. In CEAs, in particular, we often try to estimate $E[\frac{\text{effect}}{\text{cost}}]$. This is different from both $E[\frac{\text{cost}}{\text{effect}}{]}^{-1}$ and $\frac{E[\text{effect}]}{E[\text{cost}]}$ (which are also different from each other). ^[1] The goal of this post is to make this clear.

One way to simplify this is to assume that the cost is constant. So we only have uncertainty about the effect. We will also assume at first that the effect can only be one of two values, say either 1 QALY or 10 QALYs with equal probability.

Expected Value is defined as the weighted average of all possible values, where the weights are the probabilities associated with these values. In math notation, for a random variable $X$,

where $x$ are all of the possible values of $X$.^[2] For non-discrete distributions, like a normal distribution, we'll change the sum with an integral.

Coming back to the example above, we seek the expected value of effect over cost. As the cost is constant, say $C$ dollars, we only have two possible values:

In this case we do have $E[\frac{\text{effect}}{\text{cost}}]=\frac{E[\text{effect}]}{E[\text{cost}]}$, but as we'll soon see that's only because the cost is constant. What about $E[\frac{\text{cost}}{\text{effect}}]$?

which is not $\frac{1}{E[\frac{\text{effect}}{\text{cost}}]}=C\frac{2}{11}\frac{\$}{\text{QALY}}$, a smaller amount.

The point is that generally $\frac{1}{E[X]}\ne E[\frac{1}{X}].$ In fact, we always have $\frac{1}{E[X]}\le E[\frac{1}{X}]$ with equality if and only if $X$ is constant.^[3]

Another common and useful example is when $X$ is lognormally distributed with parameters $\mu ,{\sigma }^2$. That means, by definition, that  $\ln X$ is normally distributed with expected value and variance $\mu ,{\sigma }^2$ respectively. The expected value of $X$ itself is a slightly more complicated expression:

Now the fun part: $\frac{1}{X}$ is also lognormally distributed! That's because $\ln \frac{1}{X}=-\ln X$. Its parameters are $-\mu ,{\sigma }^2$ (why?) and so we get 

In fact, we see that the ratio between these values is 

1. ^{^}

   See Probability distributions of Cost-Effectiveness can be misleading for relevant discussion. There are arguably reasons to care about the two alternatives $E[\frac{\text{cost}}{\text{effect}}{]}^{-1}$ or $\frac{E[\text{effect}]}{E[\text{cost}]}$ rather than  $E[\frac{\text{effect}}{\text{cost}}]$, which are left for a future post.

2. ^{^}

   One way to imagine this is that if we sample $X$ many times we will observe each possible value $x$ roughly $P(X=x)$ of the times. So the expected value would indeed generally be approximately the average value of many independent samples.

3. ^{^}

   Due to Jensen's Inequality.