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Let $X$ follow a Beta distribution in $[0,1]$ with some parameters, and $Y$ be a non-negative random variable, independent of $X$.

Is it possible for $Y$ to have a distribution such that $Z = XY$ follows an Exponential distribution?

A bit of elaboration
By the formula for the density of the product distribution we can write

$$f_z(z) = \int_0^1\frac {1}{x}f_x(x)f_y(z/x)dx = \int_0^1\frac {dF_x(x)}{dx}\left(\frac {1}{x}f_y(z/x)\right)dx$$

$$=F_x(x)\left(\frac {1}{x}f_y(z/x)\right)|^1_0 - \int_0^1F_x(x)\left(-\frac {1}{x^2}f_y(z/x)-\frac {z}{x^3}f'_y(z/x)\right)dx$$

$$=f_y(z) + \int_0^1\frac {1}{x^2}F_x(x)\left(f_y(z/x)+\frac {z}{x}f'_y(z/x)\right)dx \tag{1}$$

To obtain the last step, I think that we have already assumed that $f_y(\cdot)$ goes to zero "fast enough".

If this is to be a density for an r.v. in $[0.\infty]$, it must be the case that

$$\int_0^{\infty}f_y(z)dz + \int_0^{\infty}\int_0^1\frac {1}{x^2}F_x(x)\left(f_y(z/x)+\frac {z}{x}f'_y(z/x)\right)dx dz = 1 \tag{2}$$

But if $f_y(\cdot)$ is a densirty of a non-negative randomm variable, it follows that the first intergal should equal unity on its own (here $z$ is just the dummy variable of intergation). So a -necessary- condition is that

$$\int_0^{\infty}\int_0^1\frac {1}{x^2}F_x(x)\left(f_y(z/x)+\frac {z}{x}f'_y(z/x)\right)dx dz =0$$

where $F_x(x)$ is the cdf of the Beta distribution.

So the question is: is there a density $f_y(\cdot)$ such that it satisfies $(2)$ and makes $(1)$ equal to $\lambda e^{-\lambda z}$ for $\lambda>0$?

I am concerned mainly with existence, I don't expect to obtain $f_y(\cdot)$.

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Based on multiple answers to Distribution of $XY$ if $X \sim$ Beta$(1,K-1)$ and $Y \sim$ chi-squared with $2K$ degrees , let $X \sim \mathrm{Beta} \left(1,K-1 \right) $ and $Y$ be chi-squared with $2K$ degrees of freedom.

Then the product is exponential with parameter $\frac{1}{2}$ .

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    $\begingroup$ (+1), and of course, $Z = \frac{2XY}{\lambda} \sim Exp(\lambda)$. $\endgroup$ – knrumsey Oct 23 '17 at 17:21
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A thought:
Since we are prescribing the distributions of $Z, X$, then $Z = XY \implies Y = Z/X$, so why not try to calculate $Y$ as a ratio distribution and see what we get?

We have

$$f_y(y) = \int_0^1 x f_z(yx)f_x(x)dx$$

We assume that $X$ is a $Beta(a,b)$ and $Z$ is an $Exp(\lambda)$ (rate parameter). So

$$f_y(y) = \frac {\lambda}{B(a,b)} \int_0^1 x \exp \{(-\lambda y)x\} x^{a-1}(1-x)^{b-1}dx$$

$$=\frac {\lambda}{B(a,b)} \int_0^1 x^a (1-x)^{b-1}\exp \{(-\lambda y)x\} dx$$

Looking up Gradshteyn & Ryzhik (2007, 7th ed.) p 347, eq. 3.383(1) and mapping coefficients, we get

$$f_y(y) = \lambda \frac {B(b,1+a)}{B(a,b)} \cdot {}_1F_{1}(1+a;1+a+b;-\lambda y)$$

where the function is Kummer's confluent hypergeometric function.

Is this a probability density over $[0,\infty]$?

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    $\begingroup$ $X$ and $Z$ and not independent, so it's difficult to see how you can justify the step at "We have...". $\endgroup$ – whuber Oct 23 '17 at 21:13
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    $\begingroup$ @whuber Oops, and I was wondering why it appears so easy... $\endgroup$ – Alecos Papadopoulos Oct 23 '17 at 21:16
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    $\begingroup$ @whuber But help me understand this: Let's assume that $Z$, $X$ are two independent r.v.'s. Then we go on and calculate the density of $Y= Z/X$ as above. Is it not still true that $XY = Z$ and so that the distribution of $XY$ will be the same as the distribution of $Z$? I have a hard time thinking otherwise. $\endgroup$ – Alecos Papadopoulos Oct 23 '17 at 21:22
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    $\begingroup$ Then X and Y would generally be dependent (because of how Y was constructed from X), and your problem statement has them be independent. $\endgroup$ – Glen_b -Reinstate Monica Oct 23 '17 at 23:57

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