Is it true that $\mathbb E[{\frac{X}{Y}]}={\frac{\mathbb E[X]}{\mathbb E[Y]}}$?

Question

If $X$ and $Y$ are both random variables, does it hold

$$\mathbb E\left[\frac{X}{Y}\right]={\frac{\mathbb E[X]}{\mathbb E[Y]}}$$

??

Answer 1

To make life simple, let $X=1$ with probability $1$.

Let $Y=2$ with probability $1/2$, and $4$ with probability $1/2$.

Then $E(Y)=3$, so $\frac{E(X)}{E(Y)}=\frac{1}{3}$.

Now we compute $E(X/Y)$, which is $E(1/Y)$. We get $\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{4}$. This is not $\frac{1}{3}$.

Truth is that almost any attempt at a counterexample works.

Answer 2

No. Take, for example, $X\equiv 1$ and $Y$ following the uniform distribution on $[0, 1]$ (or some other interval, if you want $E[X/Y]$ to be finite).

Answer 3

While there are indeed the product rule $E[XY]=E[X]E[Y]$ (this is only for $X,Y$ independent only), which make it seems like the above might work, it is however not true that $E[\frac{1}{Y}]=\frac{1}{E[Y]}$. In fact, by Jensen's inequality (assume $Y$ take value from the same sign only) we have $E[\frac{1}{Y}]\geq\frac{1}{E[Y]}$, with the equality sign only if $Y$ can only take 1 value almost surely.

Answer 4

No, This is actually a big mistake many make when thinking of averages. Seeing this through example, let $X$ represent distance traveled by car on highway and $Y$ is time traveled by car on highway, thus $E[\frac{X}{Y}]$ would be average velocity of car ride. We should not expected to get $E[\frac{X}{Y}]$ by taking our average distance traveled and dividing it by average time traveled to get average velocity. To see this more intuitively I will use fact of Law of Large Numbers that average of sample will be close to expected value in order to look in perspective of observed data averages. If you where to record data on cars on highway recording distance traveled and time to cover that distance, when using average distance traveled divided by average time to get average velocity you are ignoring how these observed points behave jointly. Some cars traveling same distance may cover these distances in less time than others, it can even be possible that cars that travel longer distances take shorter time to cover these distances than other cars traveling shorter distances. Basically, if you believe that you can just calculate both averages and divide them to get average velocity, what you are saying is if I took the data of time and distance we had and randomly assigned a time data point to a distance data point then this average velocity would still be the same for this data set since still have same average time and average distance( but this doesn't make intuitive since randomly assigning should create two completely different data sets altogether with a different average velocity for it) The point being you can't look at these points independently, can't think that $E[\frac{1}{Y}]=\frac{1}{E[Y]}$, among other things.

To learn more about mistakes in this thinking, you should check out book, Flaw of Averages http://flawofaverages.com

Note: Im sorry if this example wasn't very good, tried my best to explain in the most basic sense using a real world example. Also if there is anything wrong with this example please comment below