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I have a dart board of radius R. You hit a dart. What is the expected distance of your hit with respect to the center of the board?

Attempt:

Let $x$= distance away from the center.

$E[X]=\int_{x=0}^{R}1-P(X\leq x) dx$

And $P(X\leq x)=x^2\pi/(\pi R^2)$

Do you guys think my approach is correct?

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    $\begingroup$ Yesterday this same question was posted under a different account. It was edited a few times, until it was exactly the same letter-for-letter as this question, even with the same number of blank lines between paragraphs in the input text. About half an hour ago that question was deleted, and a minute later this one was posted. Did you temporarily lose access to your account and post under another? If so, for future reference, next time you run into a difficulty like that flag your question for moderator attention--they can probably fix the problem without deleting and reposting questions. $\endgroup$
    – David K
    Dec 11, 2016 at 17:54
  • $\begingroup$ If you are also the owner of the account 'wrek' you might still want to ask the moderators to merge the accounts in order to move your other recent question to this account. $\endgroup$
    – David K
    Dec 11, 2016 at 17:58
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    $\begingroup$ @kuku : can you help me understand why E[X] has (1-P(X<= x)) and not P(X=x) ? $\endgroup$
    – Anuj Gupta
    Aug 9, 2017 at 9:48

1 Answer 1

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Your approach is correct, although I would suggest using the notation more carefully. In particular, if you are taking the integral of something with multiple terms, put parentheses around the expression to more clearly indicate you mean the integral of the entire expression: $$E[X]=\int_0^R (1-P(X\leq x)) \,dx.$$

Note that it is not necessary to write $x=0$ at the lower end of the integral sign; the symbol $dx$ tells us that the integration is over the variable $x.$

Comparing your formula with this other answer, keeping in mind that $F_X(x)$ in that other answer is the same as your $P(X\leq x),$ we see that your formula is almost the same. In fact, a standard formula for the expectation of a random variable, when the variable takes only positive values, is

$$E[X]=\int_0^\infty (1-P(X\leq x)) \,dx.$$

In the problem you set out to solve, $1 - P(X \leq x) = 0$ whenever $x \geq R,$ so $$\int_R^\infty (1-P(X\leq x)) \,dx = 0$$

and therefore $$\int_0^\infty (1-P(X\leq x)) \,dx =\int_0^R (1-P(X\leq x)) \,dx.$$

So you are justified in taking the integral only from $0$ to $R.$

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