# Finding pdf of Quotient

#### StatStudent3

##### New Member
Let X ~ Uniform(0, 1) and Y ~ Exponential(1) be independent. Find pdf of X/Y

For Y/X case, I understand that the pdf of this is integral of |x|*f(x)*f(xz)

However, for the X/Y case, shouldn't the pdf be |y|*f(yz)*f(y) because X <= Yz for Y < 0 and X >= Yz for Y > 0 ?

--> Double Edit: My professor is correct about X/Y being the same as Y/X The reason why I'm asking this is because during the review today, my professor put the pdf formula for the Y/X case for the X/Y case - so are they same same no matter what is the numerator or denominator?

As for solving the rest of the problem, I'm fine with it. But needed to clear this particular part.

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#### StatStudent3

##### New Member
Edit: Now I'm not 100% sure...

What I realized just now is that with X/Y <= Z --> Y >= (X/z)

And now I see why this formula stays the same at integral of x * f(x) * f(x/z) (slightly, as y = x/z instead of xz)

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#### StatStudent3

##### New Member
FYI - ***SOLVED***

What I realized just now is that with X/Y <= Z --> Y >= (X/z)

And now I see why this formula stays the same at integral of x * f(x) * f(x/z) (slightly, as y = x/z instead of xz)
I just want to make sure, although I understand it, I'm not 100% - is this the correct way?

#### StatStudent3

##### New Member
I just want to make sure, although I understand it, I'm not 100% - is this the correct way?
Now that it's X/Y (flip-flopped from Y/X), do I have these steps correct...

X/Y = Z

X = Y*Z

Y = X/Z

Even though we know that Y can not be zero in this case, do you still have double integral of x * f(x) * f(x/z) for the cdf, and then single integral of the same for the pdf?

#### StatStudent3

##### New Member
Now that it's X/Y (flip-flopped from Y/X), do I have these steps correct...

X/Y = Z

X = Y*Z

Y = X/Z

Even though we know that Y can not be zero in this case, do you still have double integral of x * f(x) * f(x/z) for the cdf, and then single integral of the same for the pdf?
The test is tomorrow, and just want to make sure I have this aspect correct. There was a HW problem like this, but the TA grader marked it wrong, but didn't explain while points were taken off.

So basically, now that X/Y is flip-flopped from Y/X, even though the case now is that Y can not be equal to 0, the cdf still has (double integral)x * f(x) * f(x/z) ? In the homework, I put y instead of x, and f(yz) instead. The grader took off 2 points of the 5 worth.

#### StatStudent3

##### New Member
****FINALLY SOLVED****

I was turning the pages in my textbook for other things, and I ended up running into an example problem where finding the distribution of X/Y is the same as finding the distribution of Y/X...double integral of x * f(x) * f(xz) either way.

Could someone explain why they are the same? Otherwise, I think I should do OK if I was presented a problem with this on the test tomorrow.