3
$\begingroup$

The statement is: Let $X$ be a real-valued random variable with characteristic function $\phi_X(.)$.Let $Z=N(0,1)$ be independent of $X$.For each $\sigma>0$ the random variable $X_{\alpha}=X+\sigma Z$ has a density $f_{\alpha}$ given by,
$$f_{\alpha}(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{itx}\phi_{X}(t)e^{\frac{-\sigma^2t^2}{2}}dt$$ proof: Fix $\sigma>0$ using the independence of $X$ and $Z$ we have
$$P(X_{\sigma}{\leq\alpha})=\int_{\mathbb{R}}F_Z(\frac{\alpha-x}{\sigma})d\mu_X(x)$$ $$\int_{\mathbb{R}}\int_{-\infty}^{\frac{\alpha-x}{\sigma}}\frac{1}{\sqrt{2\pi}}e^\frac{-a^2}{a}da d\mu_X(x)$$ I have marked the step where I am getting confused. What I have understood is that $$P(X_{\sigma}\leq\alpha)=P(X+\sigma Z\leq \alpha)$$ now I am confused how we got this equals to the integral given in the proof.

$\endgroup$
3
  • $\begingroup$ You are requested not to use pictures for critical portions of the post. Please type the contents out in mathjax. $\endgroup$ Sep 26, 2022 at 14:50
  • $\begingroup$ Thanks,I will type it. $\endgroup$
    – Andyale
    Sep 26, 2022 at 14:52
  • $\begingroup$ Is there anything you can do to make your post's title useful? Like, perhaps "a question about theorem ___ in the book ___ " $\endgroup$
    – rschwieb
    Sep 26, 2022 at 15:12

1 Answer 1

4
$\begingroup$

This follows from independence and the Fubini-Tonelli theorem. To see this, let $\alpha \in \mathbb{R}$ and $A = \{(u,v) \in \mathbb{R}\times\mathbb{R} : u + \sigma v \leq \alpha \}$. Then,

\begin{align} P(X + \sigma Z \leq \alpha) &= P((X, Z) \in A)\\ &= \int_\mathbb{R} \int_\mathbb{R} 1_A(u, v)f_X(u)f_Z(v)dudv\\ &=\int_\mathbb{R} \left( \int_\mathbb{R} 1_A(u, v)f_Z(v)dv \right) f_X(u)du\\ &= \int_\mathbb{R} \left(\int_{-\infty}^{(\alpha - x)/\sigma}f_Z(v)dv \right) f_X(u)du\\ &= \int_\mathbb{R}F_Z\left(\frac{\alpha - x}{\sigma}\right)f_X(u)du \\ &= \int_\mathbb{R}F_Z\left(\frac{\alpha - x}{\sigma}\right)d\mu_X (x) \end{align} The next line then follows from the fact that $Z\sim \mathcal{N}(0, 1)$. Note that I've assumed $X$ to have a density, to make it clear where independence comes in.

$\endgroup$
2
  • $\begingroup$ @jeniffer Did this solve your problem? $\endgroup$ Sep 26, 2022 at 20:30
  • 1
    $\begingroup$ @jennifer I find that you have never voted. This site depends heavily on voting. Be reasonable, please. $\endgroup$ Oct 8, 2022 at 5:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .