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Let $X_1,X_2, . . . ,X_n$ be independent and identically distributed with $P(X_i = 1) = P(X_i = −1) = p\ $and$ P(X_i = 0) = 1 − 2p$ for all $i = 1, 2, . . . , n.$ Define

$a_n=P\bigg(  \prod_{i=1}^{\infty} X_i=1 \bigg )$, $b_n=P\bigg(  \prod_{i=1}^{\infty} X_i=-1 \bigg )$, $c_n=P\bigg(  \prod_{i=1}^{\infty} X_i=0 \bigg )$

Which of the following is true as $n$ tends to infinity?

  1. $a_n \rightarrow1/3, b_n \rightarrow1/3,c_n \rightarrow1/3$
  2. $a_n \rightarrow p, b_n \rightarrow p,c_n \rightarrow 1-2p$
  3. $a_n \rightarrow1/2, b_n \rightarrow1/2,c_n \rightarrow0$
  4. $a_n \rightarrow0, b_n \rightarrow0,c_n \rightarrow1$
in Calculus by Boss (41.9k points) | 30 views

1 Answer

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Ans will be $-B)$

$a_{n}=P\left ( \prod _{i=1}^{\infty} X_{i}=1\right )$

$=P\left ( X_{1}=1 \right )\times P\left ( X_{2}=1 \right )\times ...........P\left ( X_{n}=1 \right )$

$=p\times p\times p........\times \infty=p^{\infty}$

Now , if $p=+ve $ number  then $a_{n}\rightarrow\infty$, and if $p=fraction$, then $a_{n}\rightarrow0$, and if is $-ve$ number then $a_{n}\rightarrow-\infty$

So, it totally depends on $p$

Similarly,

$b_{n}=P\left ( \prod _{i=1}^{\infty} X_{i}=-1\right )$

$=P\left ( X_{1}=-1 \right )\times P\left ( X_{2}=-1 \right )\times ...........P\left ( X_{n}=-1 \right )$

$=p\times p\times p........\times \infty=p^{\infty}$

Same as previous

Now,

$c_{n}=P\left ( \prod _{i=1}^{\infty} X_{i}=0\right )$

$=P\left ( X_{1}=0 \right )\times P\left ( X_{2}=0 \right )\times ...........P\left ( X_{n}=0 \right )$

$=(1-2p)\times (1-2p)\times (1-2p)........\times \infty=(1-2p)^{\infty}$

Now, if $p=+ve$ integer , then value of $c_{n}$ goes to $-ve$ integer.

If $p=fraction$ , then value of $c_{n}$ goes to 0.

if $p=-ve$ integer, then value of $c_{n}$ goes to $+ve$ integer.

So, it totally depend on value of $p$
by Veteran (117k points)

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