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A random variable Y has density function f(y) = y/4 for 1<=y<3, and 0 elswhere.

The question states: Find the support of U=3Y -2, that is , find the range of real numbers u for which the density function of U is positive.

I am not sure what is the meaning of the term ‘support’ (I cannot find a definition in my text, but my understanding is that it refers to the domain or range of a function)

If this is correct, then my answer is 1<=U<7 (by substituting (U + 2)/3 into the given range of y.

But I get the density function of u to be (u +2)/36 for 1<=U<7, so the range of real numbers for which the density function of U is positive is u > -2.

Could anyone plse help clarify the term ‘support’ and whether the definition ‘the range of real numbers u for which the density function of U is positive’ is correct, and how do we obtain the support of Y.

Denson
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  • For reasons why the definition of the support you were given, is inadequate, and for a correct definition, see my comment to @heropup's answer. – Did Jul 25 '18 at 10:10

1 Answers1

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The question states the meaning of "support":

Find the support of $U = 3Y - 2$; that is, find the range of real numbers $u$ for which the density function of $U$ is positive.

The boldface part is the explanation of the meaning of "support."

You have correctly determined that the support of $U$ is $1 \le U < 7$.

heropup
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  • I was initially confused because the density function of u: (u +2)/36 is positive also outside this range (eg if u=0, or u=8), but I realize that this would not satisfy the given range of Y. Thanks very much for the clarification. – Denson Feb 26 '18 at 06:31
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    Things are slightly more complicated than that, I am afraid: first, the PDF is not unique since two nonnegative functions equal almost everywhere are equally likely candidates to be the PDF of a given absolutely continuous random variable; as a consequence, the support cannot be defined unambiguously and, as a consequence of this consequence, people often define the support of any random variable $X$, continuous or not, as the smallest closed set $C$ such that $P(X\in C)=1$. In the OP's case, this would lead to the support being $[1,7]$, not $[1,7)$. – Did Jul 25 '18 at 10:07