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## Libniz 3 years ago Let X be a discrete random variables with PMF given by : P_X(x)= x/15 ,{x=1,2,3,4,5} 0 , otherwise b)let Y=(X-3)^2, find range Y, S_Y , PMF of Y

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1. bahrom7893

@amistre64

2. Libniz

@Zarkon

3. Zarkon

where are you stuck?

4. Libniz

P_X(x)= x/15 ,{x=1,2,3,4,5} 0 , otherwise so to find range using Y y= (X-3)^2 ((x/15)-3)^2 y= (1/15 -3)^2 (-44/45)^2 to 0

5. Libniz

am I right?

6. Zarkon

no

7. Zarkon

x takes the values 1,2,3,4,5 y=(x-3)^2 plug those x values into the above equation

8. Zarkon

for the range(support)

9. Libniz

oh, it is X, not x

10. Zarkon

yes...so X takes the values 1,2,3,4,5 Y=(X-3)^2 plug those X values into the above equation :)

11. Libniz

12. Zarkon

ok...do what I wrote

13. Libniz

(1-3)^2,(2-3)^2,(3-3)^2,(4-3)^2,(5-3)^3 4,1,0,1,4

14. Zarkon

so 0,1,4

15. Zarkon

that is the range (or support)

16. Zarkon

now you can find the pmf

17. Libniz

right?

18. Libniz

so y=(x-3)^2 do we solve for x?

19. Zarkon

no

20. Zarkon

$P_{Y}(0)=P_{X}(3)$ $P_{Y}(1)=P_{X}(2)+P_{X}(4)$ $P_{Y}(4)=P_{X}(1)+P_{X}(5)$

21. Libniz

can you explain me how you got that?

22. Zarkon

$P_{Y}(1)=P(Y=1)=P(X=2\text{ or }X=4)$ $=P(X=2)+P(X=4)=P_{X}(2)+P_{X}(4)$

23. Zarkon

ok?

24. Libniz

still figuring out the last post

25. Zarkon

ok

26. Libniz

I understand you sum because it is 'or' but why is P(y=1) = P(x=2 or x=4)

27. Zarkon

$Y=(X-3)^2$ $1=(2-3)^2$ $1=(4-3)^2$ if I tell you Y=1 then either X=2 or X=4

28. Libniz

got it

29. Zarkon

good

30. Libniz

so that's just P(Y=1) do we find Y=1,2,3,4,5?

31. Zarkon

Y only takes the values 0,1,4

32. Libniz

got it , thanks; I wish you were my probability professor

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