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A. 543. The sum of the probabilities of some (finitely many) independent events is 4. Prove that with probability higher than 1/2, the number of occurred events is congruent to 0 or 3 modulo 4.

(Suggested by Endre Csóka, Budapest)

(5 points)

Deadline expired on 10 November 2011.


Solution. Let the probabilities be p1,...,pn, where p1+...+pn=4. Let xj=1 if the jth event occurs, otherwise let xj=0. Let N=x1+...+xn be the number of occuring events and let q_r = P\Big(N\equiv r\pmod4 \Big) (r=0,1,2,3). We have to show that q0+q3>q1+q2.

Consider the random complex number S=iN. This number is respectively 1, i, -1 or -i, if the remainder of N modulo 4 is 0, 1, 2, or 3.

Since

E(S)=q0+q1i+q2(-1)+q3(-i),

the statement is equivalent to


-\frac\pi4 < \arg S < \frac{3\pi}4. (1)

By the definition of S we have


S = i^N = i^{x_1+\dots+x_n} = \prod_{j=1}^n i^{x_j}
= \prod_{j=1}^n \Big((1-x_j)+x_ji\Big).

The factors on the right-hand side are independent, so


E(S) = \prod_{j=1}^n E\Big((1-x_j)+x_ji\Big)
= \prod_{j=1}^n \Big((1-p_j)+p_ji\Big).

Let \varphi=\mathrm{arg}(p_j+(1-p_j)i) (0\le\varphi_j\le\frac\pi2). To prove (1) it is sufficient to prove

 \frac54\pi < \varphi_1+\dots+\varphi_n < \frac94\pi. (2)

Leet f(t)=\arctan\frac{t}{1-t} if 0\let<1 and let f(1)=\lim_{t\to1-0}f(1)=\frac\pi2. Then \varphij=f(pj) for every j. We show that

 t \le f(t) \le \frac{49}{29}t. (3)

From this (2) immediately follows because


\varphi_1+\dots+\varphi_n = f(p_1)+\dots+f(p_n) \ge p_1+\dots+p_n = 4 >\frac54\pi,

and


\varphi_1+\dots+\varphi_n = f(p_1)+\dots+f(p_n) \le \frac{49}{29}(p_1+\dots+p_n) = \frac{49}{29}\cdot 4 < \frac94\pi.

Let g(t)=f(t)-t. For 0<t<1 we have g'(t)=f'(t)-1=\frac{2t(1-t)}{1-2t(1-t)}>0, the function g increases. Therefore, g(t)\geg(0)=0 and f(t)\get.

Now consider the function h(t)=f(t)-\frac{49}{29}t. Since h'(t)=f'(t)-\frac{49}{29}=-\frac{2(7t-2)(7t-5)}{29(1-2t(1-t))}, we have h'<0 in the intervals (0,2/7) and (5/7,1), and we have h'>0 in the interval (2/7,5/7). Therefore, h decreases in [0,2/7] and [5/7,1], and increases in [2/7,5/7]. Since h(0)=0 and h(5/7)=\arctan\frac52-\frac{49}{29}\cdot\frac57\aprox-0,0166<0, it follows that h is never positive, so h(t)\le0 and hence f(t)\le\frac{49}{29}t.


Statistics on problem A. 543.
6 students sent a solution.
5 points:Ágoston Tamás, Gyarmati Máté, Omer Cerrahoglu, Strenner Péter.
3 points:1 student.
2 points:1 student.


  • Problems in Mathematics of KöMaL, October 2011

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