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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
Umm that first line is hard to read.
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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Other answers:

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

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