The variables $X, Y, \epsilon $ are independent, with the variables $X, Y$ having a standard normal distribution $\mathcal{N}(0,1)$ and $\mathbb{P}(\epsilon = \pm 1) = \frac{1}{2}$. Does the variable $ \epsilon (X - 3Y)^2$ have a continuous distribution? If it does, find its density.
We have that:
$\mathbb{P}(\epsilon (X - 3Y)^2 < t) = \mathbb{P}((X - 3Y)^2 < t | \epsilon = 1) \cdot \mathbb{P}(\epsilon = 1) + \mathbb{P}(-(X - 3Y)^2 < t | \epsilon = -1) \cdot \mathbb{P}(\epsilon = -1)$
We know that if $A$ has a distribution given by $\mathcal{N}(a, \alpha^2)$ and $B$ has a distribution given by $\mathcal{N}(b, \beta^2)$ then:
- $A+ B$ has distribution given by $\mathcal{N}(a + b, \alpha^2 + \beta^2)$
- $uA + v$ has distribution given by $\mathcal{N}(u a + v, (u \alpha)^2)$
Therefore our $X - 3Y$ has distribution given by $\mathcal{N}(0 - 3 \cdot 0, 1 + (3 \cdot 1)^2 = \mathcal{N}(0,10) $. From that we know that:
$$\mathbb{P}((X - 3Y)^2 < t | \epsilon = 1) \cdot \mathbb{P}(\epsilon = 1) = \mathbb{P}(X - 3Y < \sqrt{t} | \epsilon = 1) \cdot \mathbb{P}(\epsilon = 1) = \frac{1}{2 \sqrt{20 \pi}} \int^{\sqrt{t}}_{- \infty} e^{- \frac{x^2}{20}} \ dx \ \ for: t \geq 0$$
and
$$\mathbb{P}(-(X - 3Y)^2 < t | \epsilon = 1) \cdot \mathbb{P}(\epsilon = 1) = \mathbb{P}(X - 3Y > \sqrt{ - t} | \epsilon = 1) \cdot \mathbb{P}(\epsilon = 1) = 1 - \mathbb{P}(X - 3Y < \sqrt{ - t} | \epsilon = 1) \cdot \mathbb{P}(\epsilon = 1) = 1 - \frac{1}{2 \sqrt{20 \pi}} \int^{\sqrt{-t}}_{- \infty} e^{- \frac{x^2}{20}} \ dx \ \ for: t < 0$$
From that we have that:
$\mathbb{P}(\epsilon (X - 3Y)^2 < t) = \frac{1}{2 \sqrt{20 \pi}} \int^{\sqrt{t}}_{- \infty} e^{- \frac{x^2}{20}} \ dx \ I_{t \geq 0} + 1 - \frac{1}{2 \sqrt{20 \pi}} \int^{\sqrt{-t}}_{-\infty} e^{- \frac{x^2}{20}} \ dx \ I_{t < 0}$
(I don't know why that line doesn't compile)
Form that we have that:
- $\displaystyle \lim_{t^{-} \to 0} \mathbb{P}(\epsilon (X - 3Y)^2 < t) = 1 - \frac{1}{2 \sqrt{20 \pi}} \int^{0}_{-\infty} e^{- \frac{x^2}{20}} \ dx $
- $\displaystyle \lim_{t^{+} \to 0} \mathbb{P}(\epsilon (X - 3Y)^2 < t) = \frac{1}{2 \sqrt{20 \pi}} \int^{0}_{-\infty} e^{- \frac{x^2}{20}} \ dx $
From that we know that it's not continous. For it to be continous $\frac{1}{2 \sqrt{20 \pi}} \int^{0}_{-\infty} e^{- \frac{x^2}{20}} \ dx = \frac{1}{2}$ and that's not the case.
Is that any correct? Are there faster ways to tell if the variable $\epsilon(X−3Y)^2$ has a continuous distribution than actually calculating it?
Avoid the mistake that:
$$\mathbb P\left [U^2\le t\right] \neq \mathbb P \left[ U \le \sqrt t\right]$$
Fix that and you will have the correct answer.
By the way in general if you have a continuous variable $U$ with pdf $f_U$ and cdf $F_U$ then $V = \epsilon U$ will also be continuous and its cdf \begin{align} F_V(v) &= \mathbb P\left[\epsilon U \le v\right]\\ &= \mathbb P\left[U \le v\right]\mathbb P\left[\epsilon = 1\right] + \mathbb P\left[U \ge -v\right]\mathbb P\left[\epsilon = -1\right]\\ &= F_U(v) \mathbb P\left[\epsilon = 1\right] + \left(1-F_U(-v)\right)\mathbb P\left[\epsilon = -1\right] \end{align}
the pdf of $V$ is:
$$f_V(v) = f_U(v) \mathbb P\left[\epsilon = 1\right] + f_U(-v)\mathbb P\left[\epsilon = -1\right]$$
Apply this for $U = (X-3Y)^2 \sim 10 \chi^2_1$. In this case
$$f_U(u) = \frac1{\sqrt{20\pi}}u^{-\frac12}e^{-\frac{u}{20}}\mathbf 1_{u\ge 0}$$