# Shopping

If Uniform Yes

#### Tsotne

Yes. I meant then you will be expecting the price to be \$N/2. So you buy exactly 2 of the items.

#### ThisGuy

If I didn't mess up the details, it works out to $$N ( 1 - \frac{\pi^2}{12})$$. Not one of your harder problems, but definitely surprising!

##### Baruch MFE Faculty
Nice work! Share the details or should I?

#### Novak

correct.
it is
$$\sum_{k=1}^\infty \large(\frac{1}{k} - \frac{1}{k+1}\right)\cdot \large(N\cdot\large(\frac{1}{k} - \frac{1}{k+1}\right)\cdot \frac{k}{2}\right)$$

one distinguishes cases depending on the number of articles bought (Pr in the first bracket) and computes the expected change in each case (the second bracket).

#### yongge

With brutal force.

For k=1,..., if x is in [N/(k+1), N/k), then
the expectation of the change is $$\int_{N/(k+1)}^{N/k} (N-kx) \frac{1}{N}dx =\frac{N}{2} \frac{1}{ k (k+1)^2}$$

Then adding them altogether, we have
$$\sum_{k=1}^\infty \frac{N}{2} \frac{1}{ k (k+1)^2}=\frac{N}{2} \sum_{k=1}^\infty \{\frac{1}{ k (k+1)} - \frac{1}{ (k+1)^2}\}$$

Ignoring the N/2 for the moment, the sum of the first term is 1 and the sum of the second term (ignoring the subtraction sign) is $$\pi^2/6-1$$
Putting these two pieces together,
we have $$\frac{N}{2} \{1 -(\pi^2/6-1)\}=N(1-\pi^2/12)$$

As other people have mentioned, for this problem, brutal force is quite easy. I'm not sure if you have a cute solution other than the brutal force.

correct.
it is
$$\sum_{k=1}^\infty \large(\frac{1}{k} - \frac{1}{k+1}\right)\cdot \large(N\cdot\large(\frac{1}{k} - \frac{1}{k+1}\right)\cdot \frac{k}{2}\right)$$

one distinguishes cases depending on the number of articles bought (Pr in the first bracket) and computes the expected change in each case (the second bracket).

#### pruse

I'm not sure if you have a cute solution other than the brutal force.

Given the presence of $$\pi$$ in the answer, do you think a "cute" solution is likely? Probably not.

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