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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?

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!

StubbornAtom
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user406509
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2 Answers2

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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.

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 $z\geq 1$.

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

Graham Kemp
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  • Thanks! Do you mind explaining how you got the integral limit of min{1, 1/z}? – user406509 Jan 13 '17 at 21:15
  • @user406509 Sure. You just partition by the cases. $$\min{1,1/z}\cdot\mathbf 1_{z>0} ~=~ \begin{cases}1 &:& 0<z<1 \ 1/z&:& 1\leqslant z\ 0 &:&\text{otherwise}\end{cases}$$ – Graham Kemp Jan 13 '17 at 23:47
  • @user406509 Also the bounds come from support for the joint pdf. $$\begin{align}\bigl{(z,\theta): 0< z\theta< 1~\wedge~ 0<\theta<1\bigr} &\equiv \bigl{(z,\theta): (0<z ~\wedge~ 0<\theta<1/z)~\wedge~0<\theta<1\bigr} \[1ex] & \equiv \bigl{(z,\theta): 0<z~\wedge~0<\theta<\min{1,1/z}\bigr}\[1ex] &\equiv\bigl{(z,\theta):0<z<1~\wedge~0<\theta<1\bigr}\cup\bigl{(z,\theta):1\leqslant z~\wedge~0<\theta<1/z\bigr}\end{align}$$ – Graham Kemp Jan 14 '17 at 00:02
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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.

Kenneth Chong
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  • Possibly easier if you deal with the cases $z \le 1$ and $z \gt 1$ seperately – Henry Jan 13 '17 at 10:06
  • Thank you so much! I tried finding the P(Z≤z) = P(X/Θ ≤ z) = P(X ≤ zΘ). I get ½ from taking the derivative of the CDF when z ≤ 1. But I'm not sure what I did wrong for the z > 1 case: I took the double integral from (0 to 1/z) and (0 to zΘ) of 1 (the joint pdf of x and Θ) dx dΘ. I got 1/(2z) and took the derivative but got -1/(2z^2). Since pdf can't be negative, do you mind explaining my mistake? – user406509 Jan 13 '17 at 21:18
  • @user406509 You made no mistake in the derivation, however a PDF is an unsigned derivative of the CDF, so you just needed to use the absolute value of the derivative. – Graham Kemp Jan 13 '17 at 23:53
  • I'm getting that $P(X/\Theta \leq z) = 1 - \frac{1}{2z}$ when z > 1. Double-check your limits of integration. The math becomes simpler if you compute $P(Z > z)$ instead, in which case, the integral becomes $$\int_0^1 \int_0^{x/z} d\theta dx$$. – Kenneth Chong Jan 14 '17 at 02:19