Marginal Density of Z = X/Θ

Question

Question: Let Θ and X be independent random variables each Uniform U(0,1) distributed.

Let Z = X/Θ

What is the marginal density of Z?

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I know that: Given Θ = θ, the density of Z is Uniform U(0, 1/θ).

However, I'm not exactly sure how to get the PDF of Z from this conditional PDF of Z given θ.

Would really appreciate any help! Thanks!

Answer 1

Start by finding $P(Z \leq z)$, for fixed $z$. You can write this probability as a double integral over the joint density between $\Theta$ and $X$. Then differentiate.

Answer 2

I know that: Given Θ = θ, the density of Z is Uniform U(0, 1/θ).

So then you know the joint distribution:$$\begin{align}f_{Z\mid \Theta}(z\mid \theta)\cdot f_{\Theta}(\theta) &= (\theta\cdot\mathbf 1_{z\in(0; 1/\theta)})\cdot(\mathbf 1_{\theta\in(0;1)})\\[1ex] &= \theta\cdot\mathbf 1_{z\in(0;\infty)}\cdot\mathbf 1_{\theta\in(0;\min\{1,1/z\})}\end{align}$$

However, I'm not exactly sure how to get the PDF of Z from this conditional PDF of Z given θ.

Just apply the Law of Total Expectation to obtain the marginal for $Z$

$$\begin{align}f_{Z}(z) &= \int_\Bbb R f_{Z\mid \Theta}(z\mid \theta)\cdot f_\Theta(\theta)\operatorname d\theta \\[1ex] &= \int_0^{\min\{1,1/z\}}\theta\cdot\mathbf 1_{z\in(0;\infty)}\operatorname d\theta \\[1ex] & = \end{align}$$

You can take it from there.

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PS: Which, by the way, is what you could obtain using vvnitram's suggestion of the Jacobian change of variables theorem.

PPS: Also note Henry's useful tip of considering the cases, $z<1$ and $z\geq 1$.

$$\begin{align}f_{Z}(z) &= \int_0^{\min\{1,1/z\}}\theta\cdot\mathbf 1_{z\in(0;\infty)}\operatorname d\theta \\[1ex] & = \mathbf 1_{z\in(0;1)}\cdot\int_0^1\theta\operatorname d \theta~+~\mathbf 1_{z\in[1;\infty)}\cdot\int_0^{1/z}\theta\operatorname d \theta \\[1ex] &= \end{align}$$