anonymous
  • anonymous
Let \[{X _{n} : n \ge1}\] be a sequence of positive independednt random variables with \[E(X _{n})=c \in(0,1)\] for each n. Let Yn=X1X2...Xn, the product of the Xi's. Use Markov's inequality to prove That Yn->0 in probability
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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blockcolder
  • blockcolder
Use Markov's Inequality with Y: \[P(|Y_n|\geq a)\leq\frac{E(Y_n)}{a}=\frac{E(X_1X_2\cdots X_n)}{a}=\frac{E(X_1)E(X_2)\cdots E(X_n)}{a}=\frac{c^n}{a}\]
anonymous
  • anonymous
I'm afraid I really need to brush up on my probability and might not be actually able to answer this question... :/ What do you mean by Y_n -> 0? Y_n is a random variable, isn't it? Surely it doesn't make sense to have a random variable tending to 0? Or does that notation mean something else?
blockcolder
  • blockcolder
Since \(c^n<1\), \(\lim_{n \to \infty} c^n=0\). Thus, by the Squeeze Theorem: \[\lim_{n \to \infty}P((Y_n-0)\geq a)=0\] which means that Y_n converges in probability to 0.

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anonymous
  • anonymous
Thanks so much, that makes a lot of sense

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