I am struggling with some basic argument in probability. Let $Z\sim N(0,1)$ and $W\sim \text{Unif}(1,2)$, and they are independent. Define $T = \frac{Z}{W}$. They we consider the conditional distribution of $T$ with respect to $W$.

Intuitively, it is obvious that $\mu(\cdot, w)$ is the same distribution as $\frac{Z}{w}$. But I am having trouble to see why. If we know the joint distribution of $T,W$ ($f(x,y)$), then we can compute the density of the conditional distribution by \begin{align*} g(x,y)=\frac{f(x,y)}{\int_{\mathbb{R}}f(x,t)dt} \end{align*}

However, we don't know the joint distribution of $T,W$ in this case. So I have no idea how can we calculate the conditional distribution formally.

I also tried to use \begin{align*} P(T\in H\,|\, W)=P\left(\frac{Z}{W}\in H\,\big|\, W\right) \end{align*} and evaluate the conditional probability at $W=w$. However, I do not know how to proceed to end up with the normal distribution.

I tried to use the uniqueness of conditional distribution, but it seems does not work. Does anyone have any idea on this?

Thanks in advance!


1 Answer 1


The distribution of $T$ conditioned on the value of $W$ is normal since we consider $W$ as a constant when evaluating the conditional distribution. Since $T=Z/W$, and $Z/a$ is normal for any nonzero value of $a$, then the distribution of $T$ given the value of $W$ must be normal.

Further, consider the CDF of $T$ conditional on the value of $W$:

$$\mathbb{P}(T \leq t | W=w) = \mathbb{P}(Z/w \leq t | W=w) = \mathbb{P}(Z \leq tw | W=w)$$

which is the CDF of a gaussian distribution given any fixed value of $w$. The existence of this distribution is proved by the Radon-Nikodym theorem.

Note that $\mathbb{P}(Z/w \leq t | W)$ is simply the random variable that takes on the value $\mathbb{P}(Z/w \leq t | W=w)$ whenever the event $\{W=w\}$ occurs.

Also note that your expression for the conditional density only holds if each of $Z$ and $W$ have a density (in this case they do).

  • $\begingroup$ This question basically is asking for a rigorous proof (involves in sigma-algebra and conditional expectation). In general, I believe we only know the existence and uniqueness of conditional distribution. However, I didn't find a theorem shows that the conditional one is the same as the one fixing the one variable . $\endgroup$
    – Zorualyh
    Sep 22, 2020 at 5:27
  • $\begingroup$ I updated with a few more details. Let me know if there is still confusion $\endgroup$
    – dmh
    Sep 22, 2020 at 13:35
  • $\begingroup$ Thanks for comment. The main confusion is that " fix $W$ as constant, the conditional distribution evaluated is indeed the conditional distribution". I have found some proof on this. They basically use Fubini theorem to show this distribution is indeed the same as the conditional distribution (Satisfying these desired properties) and conclude the statement by uniqueness of conditional distribution. $\endgroup$
    – Zorualyh
    Sep 23, 2020 at 16:26
  • $\begingroup$ There's no need to apply Fubini's theorem here, as mentioned in my answer, the existence of the conditional measure is given by the Radon Nikodym theorem. $\endgroup$
    – dmh
    Sep 23, 2020 at 16:49

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