I will mark this as homework, even though it's just general interest. The scenario is a little silly, because it's just a real world case for a theoretical problem I've been thinking about.

A man enters a hot dog eating contest every year and in the contest he has to eat 10 hotdogs in 5 minutes (yes it is different to normal hotdog contests):

As soon as the man eats his 5th hotdog (half way to finish them) the time is recorded, and when he eats his final hotdog the time is recorded again. SO we have two sets of data that can both be approximated with normal distributions, i.e. how long it take sot eat 5 hot dogs and how long to eat 10.

Call the first mean and std dev $\mu_{h}, \sigma_{h}$ (h for halfway) and the second distribution's mean and std dev $\mu_{f}, \sigma_{f}$. Both means are in minutes.

This year the man enters the contest and it is seen that he takes $\mu_{h} +\epsilon$ minutes to finish the fifth hotdog, for some $\epsilon > 0$, i.e. he is slower than usual.

My question is, what time can we expect the man two finish the contest, and how to calculate this? Thinking about it, I thought it would be best to consider the two distributions as a bivariate normal distribution. So if we have the first random variable $X$, as the time taken to get halfway, and $Y$ as the time taken to finish, I want $E(Y|X =\mu_{h} +\epsilon )$.

I don't know if this is correct thinking, please correct me if I am wrong. ALso I don't know how to go about calculating the above expected value. Any help is welcome.

  • $\begingroup$ At first I thought this was fairly straight-forward, and that you could use Poisson regression with time to 5 hotdogs as the independent variable to get the finish time 10 hotdogs given time to 5. But the more I think about it, I suspect there may be too much correlation in those two times, since the outcome is some time plus the independent variable. You might have better luck modeling the outcome as time to finish the second 5 hotdogs conditional on time to first 5 hotdogs. $\endgroup$
    – Ellie
    Commented Jun 13, 2013 at 14:23
  • $\begingroup$ The self-study tag is not just for homework but for anything that might be regarded as fairly 'standard' bookwork kind of problems. So you're probably using it correctly! $\endgroup$
    – Glen_b
    Commented Jun 14, 2013 at 1:59

2 Answers 2


Just a note on calculating these sorts of conditional expectations.

I am going to use a different notation from you, because I've dealt with this problem before (in two quite different contexts) and I want to use the notation I'm used to playing with it in (where we're not even dealing with time, but a different thing that accumulates).

Call the first 5 hotdogs 'task 1' and the next 5 hotdogs 'task 2'.

Let $X$ be the time to finish the first task, and $Y$ the time to finish the second task (so $X+Y$ is the time to finish everything).

After the first task, we observe $X=x$. We want $\text{E}(X+Y|X=x)$.

$$\text{E}(X+Y|X=x) = \text{E}(X|X=x) + \text{E}(Y|X=x) = x + \text{E}(Y|X=x)$$

Now, what do we do with $\text{E}(Y|X=x)$?

If $X$ and $Y$ are independent, we can just put $\mu_Y$; if they're nearly independent this will be a good approximation.

(which turns out to be a surprisingly good approximation in one of those tasks)

More generally, you could do something like the conditional expectation in a bivariate normal (since you mention two normals):

$$\text{E}(Y|X=x) = \mu_Y + \text{Cov}(X,Y) \text{Var}(X)^{-1} (x-\mu_X) $$

(If this looks like regression, that's no accident.)

In your case, the final term in parentheses would be $\epsilon$.

So you end up for the total time with $x + \mu_Y + \gamma(x-\mu_X)$ where $\gamma$ is essentially a population regression coefficient.

So to predict what happens to the expectation in the second task given what happened in the first, you need to know about how they co-vary relative to the variance of the first task. (Indeed, this hints that if you don't know these quantities, you would estimate the relationship from data by regression. However, if you already know/can assume either the correlation or covariance, you can write down the relevant equation.)

(I've glossed over some of the assumptions involved here.)

  • $\begingroup$ WHen you write the equation for $E(Y|X=x)$ shouldn't the multiplier be a correlation coefficient? i.e. coveriance divided by the two means? $\endgroup$
    – Solitude
    Commented Jun 14, 2013 at 14:20
  • $\begingroup$ i) no, it shouldn't be a correlation coefficient; it has to be in the units of Y divided by units of X, not scale-free and (ii) covariance divided by two means is not a correlation; it's covariance divided by two sd's. $\endgroup$
    – Glen_b
    Commented Jun 15, 2013 at 0:04

If you are happy with the bivariate normal assumption (and it seems a reasonable approximation to me), then the expected value of Y given X is just the prediction from a linear regression model. The standard formulas for linear regression can be derived either by assuming the X values are fixed, or that X and Y are bivariate normal, either way the formulas for the regression coefficients are the same.

So if you have data from the past you can just put it into a regression routine and make a prediction.

If you don't have data, but rather have the parameters of the bivariate normal (2 means, 2 variances (or standard deviations), and one covariance (or correlation)) you can plug those into the formulas and get the regression equation to predict with.


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