## Libniz 3 years ago 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)

1. Libniz

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

2. wio

Umm that first line is hard to read.

3. Libniz

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

4. Libniz

c)|dw:1348940632883:dw|

5. Libniz

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

6. Libniz

@zarkon

7. Zarkon

8. Libniz

oh, sqrt[0.5]

9. Zarkon

correct

10. Zarkon

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

11. Zarkon

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

12. Libniz

what does the comma mean P(X=1,X≥0)?

13. Zarkon

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

14. Libniz

product of the two then ?

15. Zarkon

no...they are not independent

16. Zarkon

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

17. Libniz

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

18. Zarkon

$x\ge 0\Rightarrow x\in\{0,1\}$ $\{1\}\cap\{0,1\}=\{1\}$

19. Libniz

$\frac{P(X=1)}{P(X\ge 0)}$ 0.25 ---------- = 1/3 0.75

20. Zarkon

yep

21. Libniz

you are truly a great teacher; thanks sir @Zarkon