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     \mathrm{Let\ \xi_n \ , for\ n \in \ \mathbb{N}_0 \, be\ a\ collection\ of\ random\ variables\ such\ that\ each\ \xi_n\ takes\ values\ in\ the\ set\ \{3,8,11\}}  \mathrm{with\ the\ distribution\ given\ by}:

    P_\xi_n (3) = 1 - \dfrac{3}{n} - \dfrac{5}{n^2} , P_\xi_n (8) = \dfrac{3}{n} , P_\xi_n (11) = \dfrac{5}{n^2}

     \mathrm{Prove\ that\ \xi_n \rightarrow\ 3\ as\ n\ \rightarrow\ \infty}


    I realise it's 2:40 am, and I've been working since midday, so a lack of sleep is probably hindering my ability to tackle this question.
    But even since this evening, I haven't had a clue.

    I have my notes in front of me, I have tried every avenue, from the Law of Large Numbers, to Chebyshev's Inequality, and I just cannot seem to find a starting point.

    I would post workings, but I haven't found a way to start this.
    I'm not looking for an answer, merely a push in the right direction, so if anyone could help, I would really appreciate it
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    Starting point is to note that the statement is obvious. Let n tend to infinity for each expression you have there, clearly P(8) and P(11) are tending to 0 and P(3) tending to 1 just taking normal limits.

    You can now use that information to come up with a more formal epsilon proof.
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    (Original post by Swayum)
    Starting point is to note that the statement is obvious. Let n tend to infinity for each expression you have there, clearly P(8) and P(11) are tending to 0 and P(3) tending to 1 just taking normal limits.

    You can now use that information to come up with a more formal epsilon proof.

    Okay, I have spent this week really trying to do it, but have got nowhere.

    There was a part i) to this question:

     \mathrm{ Let\ \xi\ be\ a\ random\ variable\ on\ a\ discrete\ probability\ space\ (\Omega, \mathcal{A}, P)\ and\ let\ \epsilon\ >\ \epsilon_0}

     \mathrm{ Show\ that\ P(| \xi - a| \geq \epsilon) \leq P(|\xi - a| \geq \epsilon_0)\ for\ real\ and\ positive\ a.}

     \mathrm{ Why\ is\ this\ useful\ when\ checking\ stochastic\ convergence?}

    Now, this is how I answered it:

     P(| \xi - a| \geq \epsilon) \leq \dfrac{\mathrm{Var(\xi)}}{ \epsilon^2}

     \dfrac{\mathrm{Var(\xi)}}{ \epsilon^2} < \dfrac{\mathrm{Var(\xi)}}{ \epsilon_0^2}

     P(| \xi - a| \geq \epsilon) \leq \dfrac{\mathrm{Var(\xi)}}{ \epsilon^2} < \dfrac{\mathrm{Var(\xi)}}{ \epsilon_0^2} \geq P(| \xi - a| \geq \epsilon_0)

     \dfrac{\mathrm{Var(\xi)}}{ \epsilon_0^2} \geq P(| \xi - a| \geq \epsilon_0)

     \dfrac{\mathrm{Var(\xi)}}{ \epsilon_0^2} > \dfrac{\mathrm{Var(\xi)}}{ \epsilon}

     \Rightarrow \dfrac{\mathrm{Var(\xi)}}{ \epsilon^2} \leq  P(| \xi - a| \geq \epsilon_0)

     \Rightarrow P(| \xi - a| \geq \epsilon) \leq P(| \xi - a| \geq \epsilon_0)



     \mathrm{ By\ definition\ (given\ by\ lecturer):}

     \mathrm{ Let\ \xi_n\ be\ a\ collection\ of\ random\ variables\ we\ say\ that\ \xi_n\ converges\ stochastically\ to\ a\ real\ number\ as\ n \rightarrow\ \infty\ if}  \mathrm{ \forall\ \epsilon\ >\ 0}

     \displaystyle\lim_{n\to \infty}  P(| \xi_n - a| \geq \epsilon)

     \mathrm{ As\ n \rightarrow \infty\ then\ \dfrac{\mathrm{Var(\xi)}}{n \epsilon_0^2}} \rightarrow 0

     \mathrm{ But\ \dfrac{\mathrm{Var(\xi)}}{n \epsilon_0^2} \geq\ \dfrac{\mathrm{Var(\xi)}}{n \epsilon^2} \geq P(| \xi_n - a| \geq \epsilon) > 0}

     \mathrm{ So\ by\ the\ squeezing\ principle\, we\ see\ that\ as\ n \rightarrow \infty\ the\ conditions\ hold\ so\ P(| \xi_n - a| \geq \epsilon) \rightarrow\ 0\ as\ n \rightarrow \infty}

    I emailed my lecturer about part ii), and got this as a response:

     \mathrm{ Use\ part\ (i)\ to\ ensure\ that\ you\ need\ only\ check\ things\ for\ epsilon\ smaller\ than\ some\ particular\ value}
     \mathrm{ Choose\ this\ value\ so\ that\ all\ the\ attention\ is\ on\ \xi_n\ being\ 8\ or\ 11.}

    I am really sorry for such a long post, but I wanted to show exactly what I've done, and what it is I'm not understanding.
    If I calculated  \mathbb{E}(\xi_n) then I can get it so that:

     3 \times (1 - \dfrac{3}{n} - \dfrac{5}{n^2}) + 8 \times (\dfrac{3}{n}) + 11 \times (\dfrac{5}{n^2}) = 3\ as\ n \rightarrow \infty}

    However, that is not going the way that is suggested, and even then, I don't know what to do with that

    I really am stuck on this
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     \mathrm{ A\ sequence\ \xi_n\ converges\ in\ probability\ to\ \xi\ if\ for\ every\ \epsilon\ > 0}

     P (|\xi_n - \xi| < \epsilon) \rightarrow 1\ as\ n\ \rightarrow \infty}

     \mathrm{ As\ n \rightarrow \infty;}

 P_\xi_n (3) \rightarrow 1

 P_\xi_n (8) \rightarrow 0  

 P_\xi_n (11) \rightarrow 0

    Am I even close?

    I'm planning on going in to uni tomorrow. I went in today to see him, but he'd gone home, and if he isn't in tomorrow, as it is due in monday, I don't know how I'll do it
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    Sorry, I'm not very good at this stuff, hopefully someone else can help.
 
 
 
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