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A certain country removed the $1$ cent and $2$ cent coin, and therefore was intended that every cash transaction be rounded to the nearest $5$ cent (note there are $100$ cents in every dollar.). Assuming that the number of cents in each transaction is uniformly distributed.
Amounts to be paid are rounded to the nearest $5$ cents when paid in cash. If paid electronically via credit card, the exact amount is charged.
You can abuse this system by paying for transactions in cash when the amount will be rounded down, and paying electronically (credit card or EFTPOS) when the amount would have been rounded up. Let $Y$ be the amount of money you save by using this approach, in a transaction that would be rounded to the nearest $5$ cents if paid for in cash.

It then asks for the PDF of the r.v. $Y$. I had this:

$$f_{Y}(y) = \begin{cases} \frac{3}{7} \qquad y=0 \\ \frac{2}{7} \qquad y=1\\ \frac{2}{7}\qquad y=2.\end{cases}$$

However, the solution says this:
$$f_{Y}(y) = \begin{cases}\frac{3}{5}\qquad y=0 \\ \frac{1}{5}\qquad y=1 \\ \frac{1}{5} \qquad y=2 \\ 0 \qquad \text{otherwise.}\end{cases}$$

I don't understand how it was constructed, as since the transaction only occurs when we know the cost will round down, I don't see why there would be a $\frac{3}{5}$ for example, in the PDF.

This is how I found mine: I listed out all the "possible" transaction value that could happen for any 10 cent interval:
So transactions ending in
$00$,$01$,$02$,$03$,$04$,$05$,$06$,$07$,$08$,$09$,$10$ (the $10$ represents a transacion like $\$ 1.10$ and the $00$ represents a transcation like $\$1.00$).
So for any cash transaction, I'd cross out the ones that would make me round up, and we're left with:
$00$,$01$,$02$,$05$,$06$,$07$,$10$ So there are seven possibilities where I'd make a cash transaction.
$Y=0$ means I gain nothing, so that'd be at $00$,$05$,$10$, making it have $\frac{3}{7}$ probability. This was done similarly for the other probabilities. Basically, I don't see how the answer can have denominators of $5$.
Thanks for any help

Natash1
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1 Answers1

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The complete set of options are actually only prices from 0 to 4 cents above a round nickel value. Demonstration of that and determination of the benefit:

  • 0 cents above (e.g., $\$1.00$): No savings to use cash

  • 1 cent above (e.g., $\$1.01$): 1 cent saving to use cash (would be $1.00)

  • 2 cent above (e.g., $\$1.02$): 2 cent saving to use cash (would be $1.00)

  • 3 cent above (e.g., $\$1.03$): You wouldn't use cash (since it would round up) (so a 0 cent saving)

  • 4 cent above (e.g., $\$1.04$): You wouldn't use cash (since it would round up) (so a 0 cent saving)

(When you get to $\$1.05$, you're back to the start again, at 0 cents above)

So three out of the 5 are 0 cents
One out of the 5 is 1 cent savings
One out of the 5 is 2 cent savings And no way to save a different amount

And so when written properly, that matches the valid answer.

So what went wrong in your work?

  • Choosing a 10 cent range instead of 5... it wouldn't have ended up directly hurting without the other errors, you just would have gotten reduceable fractions. However, using larger values typically does open the door to more mistakes :-)
  • By doing 0.00 AND 0.10, you duplicate that value... basically counting a certain result an extra time. It'd be like counting 1990 and 2000 as part of the 90s. Not the first to go in that direction, it is important to be particularly careful not to double-counting certain values of sets errantly...
  • Those values that would be rounded up (03, 04, 08, 09) may not be any benefit to you, but they're still valid possibilities in the set... so you can't throw them out (just as you properly didn't throw out .00) It is indeed one of the possible results for you when you walk in the store, so you can't neglect it.

So if you'd done .00 - .09, and not crossed any off, you'd have gotten
$$f_{Y}(y) = \begin{cases} \frac{6}{10} \qquad y=0 \\ \frac{2}{10} \qquad y=1\\ \frac{2}{10}\qquad y=2.\end{cases}$$

And indeed those fractions reduce to the correct answers (and adding the y=otherwise is just for full completeness)

Hope that helps, keep with it, seems like you're not too far off :-)

JeopardyTempest
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