Possible values of the probability density function (1/a)e^-((y-b)/a)?

What are the possible values of the pdf (1/a)e^-((y-b)/a)? This is the pdf of y=ax+b (x>0, b is some number), and x is exponentially 1 distributed (e^(-x)). Also, how do I find these possible values? Thank you.

Allie Z said:

What are the possible values of the pdf (1/a)e^-((y-b)/a)? This is the pdf of y=ax+b (x>0, b is some number), and x is exponentially 1 distributed (e^(-x)). Also, how do I find these possible values? Thank you.

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Are you asking for the range of the function f(x) = (1/a)e^-((y-b)/a)? Or for something else?

Possible values/support of pdf of y

Dr.Peterson said:

Are you asking for the range of the function f(x) = (1/a)e^-((y-b)/a)? Or for something else?

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I'm asking for the possible values, that is, the support of that function. That is, for example, if the probability mass function (pmf) of x is fx(k) = 1/4 for k= -1, 1/4 for k = 1, 1/2 for k = 2, and 0 otherwise, then these k values are the support/possible values of fx(k). I'm asking for the possible value/support range of a pdf.

on support again

Allie Z said:

I'm asking for the possible values, that is, the support of that function. That is, for example, if the probability mass function (pmf) of x is fx(k) = 1/4 for k= -1, 1/4 for k = 1, 1/2 for k = 2, and 0 otherwise, then these k values are the support/possible values of fx(k). I'm asking for the possible value/support range of a pdf.

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I forgot to add that the support is the range of values for which the function is positive. I figured out that in this case, the support is y > b because if x = 0, then y = b (based on y=ax+b), and if the support of the pdf of x is x>0, then the support of the pdf of y needs to be >b. I wonder though whether there is an easy way to find the support of pdfs with for example y=e^(-x) (and the pdf of x being ((1/sqrt(2pi))e^(-t^2/2), and the pdf of y being ((1/sqrt(2pi))e^((-ln^(2)(y)/(2)))(1/y).

Allie Z said:

I'm asking for the possible values, that is, the support of that function. That is, for example, if the probability mass function (pmf) of x is fx(k) = 1/4 for k= -1, 1/4 for k = 1, 1/2 for k = 2, and 0 otherwise, then these k values are the support/possible values of fx(k). I'm asking for the possible value/support range of a pdf.

Click to expand...

Allie Z said:

I forgot to add that the support is the range of values for which the function is positive. I figured out that in this case, the support is y > b because if x = 0, then y = b (based on y=ax+b), and if the support of the pdf of x is x>0, then the support of the pdf of y needs to be >b. I wonder though whether there is an easy way to find the support of pdfs with for example y=e^(-x) (and the pdf of x being ((1/sqrt(2pi))e^(-t^2/2), and the pdf of y being ((1/sqrt(2pi))e^((-ln^(2)(y)/(2)))(1/y).

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So, when you asked about "possible values of the pdf", you were really asking for "possible values" of the random variable, in the sense of values of y for which the pdf is non-zero? Can you state the exact wording of the problem you are working on, so we can be sure of it?

I may be missing something, but it seems to me that the function (1/a)e^{-(y-b)/a} that you asked about is zero for all real numbers, so its support is R.

I don't understand what you are saying about x>0, which shouldn't be relevant to the support of the pdf.

Re:

You're correct, I asked about the ""possible values" of the random variable, in the sense of values of y for which the pdf is non-zero". I'll try to find the exact wording of the question and post here again. Thank you.