CLT question, the density function of the horizontal deviation from shot arrow to center is given…

Question

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 $X_i = 1,-1$.

Answer 1

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 $100(1-(2p-1)^2)$.

Answer 2

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 $L$

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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.