Question

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)

Answer 1

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

Answer 2

Umm that first line is hard to read.

Answer 3

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

Answer 4

c)

Answer 5

d) P[x=0]+P[x=1]

Answer 6

@zarkon

Answer 7

your final answer for b is not correct

Answer 8

oh, sqrt[0.5]

Answer 9

correct

Answer 10

d)\[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}\]

Answer 11

\[P(X=1| X\ge0)=\frac{P(X=1,X\ge 0)}{P(X\ge 0)}=\frac{P(X=1)}{P(X\ge 0)}\]

Answer 12

what does the comma mean P(X=1,X≥0)?

Answer 13

\[P(X=1 \text{ and } X\ge0)\]

Answer 14

product of the two then ?

Answer 15

no...they are not independent

Answer 16

\[P(X=1 \text{ and } X\ge0)=P(X=1)\]

Answer 17

oh, "and "means it has to satisfy both condition. x=1 is inside x>=0 ; so we choose x=1

Answer 18

\[x\ge 0\Rightarrow x\in\{0,1\}\]

\[\{1\}\cap\{0,1\}=\{1\}\]

Answer 19

\[\frac{P(X=1)}{P(X\ge 0)}\]

0.25
---------- = 1/3
0.75

Answer 20

yep

Answer 21

you are truly a great teacher; thanks sir
@Zarkon