Libniz
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)
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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
wio
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Umm that first line is hard to read.
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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
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c)|dw:1348940632883:dw|
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d) P[x=0]+P[x=1]
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@zarkon
Zarkon
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your final answer for b is not correct
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oh, sqrt[0.5]
Zarkon
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correct
Zarkon
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d)\[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}\]
Zarkon
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\[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}=\frac{P(X=1)}{P(X\ge 0)}\]
Libniz
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what does the comma mean P(X=1,X≥0)?
Zarkon
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\[P(X=1 \text{ and } X\ge0)\]
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product of the two then ?
Zarkon
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no...they are not independent
Zarkon
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\[P(X=1 \text{ and } X\ge0)=P(X=1)\]
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oh, "and "means it has to satisfy both condition. x=1 is inside x>=0 ; so we choose x=1
Zarkon
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\[x\ge 0\Rightarrow x\in\{0,1\}\]
\[\{1\}\cap\{0,1\}=\{1\}\]
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\[\frac{P(X=1)}{P(X\ge 0)}\]
0.25
---------- = 1/3
0.75
Zarkon
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yep
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you are truly a great teacher; thanks sir
@Zarkon