Question

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

Answer 1

@amistre64

Answer 2

@Zarkon

Answer 3

where are you stuck?

Answer 4

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

Answer 5

am I right?

Answer 6

no

Answer 7

x takes the values 1,2,3,4,5

y=(x-3)^2

plug those x values into the above equation

Answer 8

for the range(support)

Answer 9

oh, it is X, not x

Answer 10

yes...so

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

:)

Answer 11

ok...do what I wrote

Answer 12

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

Answer 13

so 0,1,4

Answer 14

that is the range (or support)

Answer 15

now you can find the pmf

Answer 16

right?

Answer 17

so

y=(x-3)^2

do we solve for x?

Answer 18

no

Answer 19

\[P_{Y}(0)=P_{X}(3)\]
\[P_{Y}(1)=P_{X}(2)+P_{X}(4)\]
\[P_{Y}(4)=P_{X}(1)+P_{X}(5)\]

Answer 20

can you explain me how you got that?

Answer 21

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

Answer 22

ok?

Answer 23

still figuring out the last post

Answer 24

ok

Answer 25

I understand you sum because it is 'or' but

why is P(y=1) = P(x=2 or x=4)

Answer 26

\[Y=(X-3)^2\]

\[1=(2-3)^2\]

\[1=(4-3)^2\]

if I tell you Y=1 then either X=2 or X=4

Answer 27

got it

Answer 28

good

Answer 29

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

Answer 30

Y only takes the values 0,1,4

Answer 31

got it , thanks;

I wish you were my probability professor