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The horizontal deviation from shot arrow to the center of the target is given as : $$ \varphi (x)= \begin{cases} 1- |x| , x\in (-1,1) \\ 0 , \text{ otherwise. } \end{cases}$$ If the horizontal deviation from shot arrow to center is within the set $$(-\frac{1}{2}, \frac{1}{2})\cup (0,\frac{2}{3})$$ the shooter gets one point, otherwise he loses one point. What is the probability that in $100$ shots the shooter claimed less than $55$ points?

The assignment is done like $S_n=\sum{X_i}$ where $$X_i =\begin{cases} 1, \text{(success, meaning the deviation is within the set)} \\ -1, \text{the deviation is within the set $(\frac{2}{3},1)$} \end{cases}$$ Then using the CLT..

I am having trouble finding the probabilities of $X_i = 1,-1$.

Jerry West
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2 Answers2

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You have success when the deviation is in $(-1/2,2/3)$ and failure when it is in $(-1,-1/2) \cup (2/3,1)$. Accordingly, you need to compute the probability of success $p$ by doing an integral (or by drawing a figure and computing the area of a trapezoid by geometry). Once you know that, you can compute the mean and variance of the $X_i$; supposing the probability of success is $p$, these are $2p-1$ and $1-(2p-1)^2$, respectively. Then $\sum_{i=1}^{100} X_i$ is approximately normally distributed with mean $100(2p-1)$ and variance $100(1-(2p-1)^2)$.

Ian
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Define iid random variables $B_1, B_2, ...$ s.t. $B_i = 1_{A_i}$ where $A_i$ is the event of success. Then

$B_i \sim Bernoulli(p)$ where $p: = P(B_i = 1) = P(A_i)$

Define iid random variables $L_1, L_2, ...$ s.t. $L_i = 1_{A_i^C}$

$L_i \sim Bernoulli(1-p)$ where $1-p: = P(L_i = 1) = P(A_i^C)$

Observe that:

  1. $B := B_1 + ... + B_{100} \sim Binomial(100,p)$

  2. $L := L_1 + ... + L_{100} \sim Binomial(100,1-p)$

  3. $X_i = B_i - L_i$

  4. $X := X_1 + ... + X_{100} = (B_1 + ... + B_{100}) - (L_1 + ... + L_{100}) = B - L$

  5. $(X_i = 1) = (B_i = 1) = (L_i = 0) = (B_i = 1, L_i = 0)$

  6. $(X_i = -1) = (B_i = 0) = (L_i = 1) = (B_i = 0, L_i = 1)$

Now, let A be the area of success. Define iid random variables $Y_1, Y_2, ...$ with pdf $f_{Y_i}(y_i) = \varphi(y_i)$. Then, we have

$$p = P(y_i \in A) = \int_A f_{Y_i}(y_i) dy_i$$

Finally, we have (assuming I am using wolfram alpha right)

$$P(X \le 55) = P(B - L \le 55) = \sum_{b=55}^{100} \ \sum_{l=b-55}^{100} f_{B,L}(b,l)$$

where $f_{B,L}(b,l)$ is the joint distribution of $B$ and $L$

CLT allows you to approximate the probability:

$$\because \lim P(a_1 \le \frac{\frac{X_1 + ... + X_{n}}{n} - \mu}{\sigma/\sqrt{n}} \le a_2) = \int_{a_1}^{a_2} f_Z(z) dz$$

where $Z$ is standard normal, $a_1, a_2 \in \mathbb R$, $X_i$'s are iid, $\mu = E[X_i], \sigma = Var[X_i]$

and $\because$ the $B_i - L_i$'s are independent (I think),

we have:

$$P(0 \le X_1 + ... + X_{100} \le 55)$$

$$ = P(\frac{\frac{0}{100} - \mu}{1} \le \frac{\frac{X_1 + ... + X_{100}}{100} - \mu}{1} \le \frac{\frac{55}{100} - \mu}{1})$$

$$\approx \int_{\frac{\frac{0}{100} - \mu}{1}}^{\frac{\frac{55}{100} - \mu}{1}} f_Z(z) dz = N(\frac{\frac{55}{100} - \mu}{1}) - N(\frac{\frac{0}{100} - \mu}{1})$$

where $N(\cdot)$ denotes standard normal cdf.

Community
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BCLC
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    You need to adjust for the fact that failures are counted as $-1$. – Ian Jan 04 '16 at 21:44
  • @Ian Oh shoot. Thanks. Going to edit – BCLC Jan 04 '16 at 21:44
  • Who wrongfully upvoted my comment? :P – BCLC Jan 04 '16 at 22:21
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    Each $X_i=B_i-L_i$ seems to be such that $P(X_i=1)=p^2$, $P(X_i=-1)=(1-p)^2$ and $P(X_i=0)=2p(1-p)$. This is not what the exercise is about. – Did Jan 04 '16 at 22:32
  • @Did Thanks. Edited. How about now? – BCLC Jan 05 '16 at 00:12
  • Now the model is made overly complicated without need but it probably corresponds to the setting (eventhough the mention of a joint distribution of (B,L) is merely a joke). Next, the CLT: the whole part after "we have" is a complete mystery. – Did Jan 05 '16 at 00:38
  • @Ian Done. How is it now? – BCLC Jan 05 '16 at 00:41