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Libniz Group Title

P(X=-1)= P(x=+1)=1/2P(X=0) a) compute P(X=0) b)compute Standand deviation funcion of x c)draw cumulative distributive function of x d) compute P(x=1| x>=0)

  • one year ago
  • one year ago

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  1. Libniz Group Title
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    P[x=-1]+P(x=1)+P[x=0]=1 P[x=-1]+P[x=-1]=P[x=0] 2P[x=0]=1 P[x=0]=0.5 P[x=1]=P[x=-1]=.25

    • one year ago
  2. wio Group Title
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    Umm that first line is hard to read.

    • one year ago
  3. Libniz Group Title
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    b) standard deviations of X E[x]=.25(1)+.25(-1)+0(.5)=0 (E[x])^2=0 E[x^2]=.5 var[x]= .5-0=.5 standard deviation=.5

    • one year ago
  4. Libniz Group Title
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    c)|dw:1348940632883:dw|

    • one year ago
  5. Libniz Group Title
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    d) P[x=0]+P[x=1]

    • one year ago
  6. Libniz Group Title
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    @zarkon

    • one year ago
  7. Zarkon Group Title
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    your final answer for b is not correct

    • one year ago
  8. Libniz Group Title
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    oh, sqrt[0.5]

    • one year ago
  9. Zarkon Group Title
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    correct

    • one year ago
  10. Zarkon Group Title
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    d)\[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}\]

    • one year ago
  11. Zarkon Group Title
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    \[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}=\frac{P(X=1)}{P(X\ge 0)}\]

    • one year ago
  12. Libniz Group Title
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    what does the comma mean P(X=1,X≥0)?

    • one year ago
  13. Zarkon Group Title
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    \[P(X=1 \text{ and } X\ge0)\]

    • one year ago
  14. Libniz Group Title
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    product of the two then ?

    • one year ago
  15. Zarkon Group Title
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    no...they are not independent

    • one year ago
  16. Zarkon Group Title
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    \[P(X=1 \text{ and } X\ge0)=P(X=1)\]

    • one year ago
  17. Libniz Group Title
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    oh, "and "means it has to satisfy both condition. x=1 is inside x>=0 ; so we choose x=1

    • one year ago
  18. Zarkon Group Title
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    \[x\ge 0\Rightarrow x\in\{0,1\}\] \[\{1\}\cap\{0,1\}=\{1\}\]

    • one year ago
  19. Libniz Group Title
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    \[\frac{P(X=1)}{P(X\ge 0)}\] 0.25 ---------- = 1/3 0.75

    • one year ago
  20. Zarkon Group Title
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    yep

    • one year ago
  21. Libniz Group Title
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    you are truly a great teacher; thanks sir @Zarkon

    • one year ago
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