# The pdf of a standard uniform random variable divided by constant [closed]

For a random variable $$\frac{U}{a}$$ where $$U$$ is a standard uniform random variable, I'm trying to determine the pdf. I'm not so sure what I'm getting is correct as I'm getting some funny results down the line. Can someone please weigh in on where I'm off, if anywhere?

For a random variable $$X$$, the random variable $$Y=aX+b$$ has pdf $$\frac{1}{|a|}f_X(\frac{y-b}{a})$$. So, recognizing $$\frac{U}{a}$$ as $$Y=\frac{1}{a}U+0$$, then the pdf should be $$af_U(ny)$$.

Because the pdf of $$U$$ is $$f_U(y)=1$$, then $$af_U(ny)=a$$, ie the pdf of a uniform random variable divided by a constant is simply that constant.

Any input?

• Given $U$ follows standard uniform distribution, your conclusion is right. Commented Dec 4, 2018 at 0:24
• Support of $U$ is $(0,1).$ Should mention support of $Y.$ Commented Dec 4, 2018 at 7:37
• You have yet to present any valid PDF in this post: every one you describe has a divergent integral. The reason is that you have neglected to mention that the uniform PDF is zero outside an interval. Although this might seem like a trivial technical point, neglecting it often is at the root of "funny results down the line."
– whuber
Commented Dec 4, 2018 at 15:55
• What is $n$ in $af_U(ny)$?? Commented Dec 15, 2018 at 14:59
• @DilipSarwate The correct notation for this question amounts to asking what a PDF is for a uniform distribution on min to max. As it stands, this question is undefined as per whuber's comment, and the answer provided uses the same bad notation. Unfortunately, posting what an answer looks like, is the same as posting what a uniform distribution actually is.
– Carl
Commented Dec 15, 2018 at 20:43

If we write $$V = U/a$$ then, assuming $$a > 0$$, $$f_{V}(v) = af_{U}(av) = aI_{[0, 1]}(av) = aI_{[0, 1/a]}(v),$$ which is a proper density since $$\int_0^{1/a}adv = a[1/a - 0] = 1.$$
If $$a < 0,$$
then we get, in a similar fashion, that $$f_V(v) = |a|I_{[1/a, 0]}(v).$$