Libniz
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=(X3)^2, find range Y, S_Y , PMF of Y



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bahrom7893
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@amistre64

Libniz
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@Zarkon

Zarkon
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where are you stuck?

Libniz
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P_X(x)= x/15 ,{x=1,2,3,4,5}
0 , otherwise
so to find range using Y
y= (X3)^2
((x/15)3)^2
y= (1/15 3)^2
(44/45)^2
to 0

Libniz
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am I right?

Zarkon
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no

Zarkon
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x takes the values 1,2,3,4,5
y=(x3)^2
plug those x values into the above equation

Zarkon
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for the range(support)

Libniz
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oh, it is X, not x

Zarkon
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yes...so
X takes the values 1,2,3,4,5 Y=(X3)^2 plug those X values into the above equation
:)

Libniz
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Zarkon
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ok...do what I wrote

Libniz
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(13)^2,(23)^2,(33)^2,(43)^2,(53)^3
4,1,0,1,4

Zarkon
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so 0,1,4

Zarkon
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that is the range (or support)

Zarkon
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now you can find the pmf

Libniz
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right?

Libniz
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so
y=(x3)^2
do we solve for x?

Zarkon
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no

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

Libniz
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can you explain me how you got that?

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

Zarkon
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ok?

Libniz
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still figuring out the last post

Zarkon
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ok

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

Zarkon
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\[Y=(X3)^2\]
\[1=(23)^2\]
\[1=(43)^2\]
if I tell you Y=1 then either X=2 or X=4

Libniz
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got it

Zarkon
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good

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

Zarkon
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Y only takes the values 0,1,4

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