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===The case of Sums of independent Random Variables=== | |||
Let <math>X</math> be a Poisson distributed r.v. with parameter <math>\lambda > 0</math>. We already know that for all <math>\xi\in \R</math>, we then have <math>\E\left[e^{i\xi X}\right]=\exp(-\lambda(1-e^{i\xi}))</math>. Let <math>X_1,...,X_n</math> be <math>n</math> independent r.v.'s for <math>n\in\N</math> with <math>X_j</math> being a Poisson distributed r.v. with parameter <math>\lambda_j > 0</math> for all <math>1\leq j\leq n</math>. Now let <math>S_n=\sum_{j=1}^nX_j</math>. We want to figure out what the law of <math>S_n</math> is. We have | |||
<math display="block"> | |||
\begin{align*} | |||
\E\left[e^{i\xi S_n}\right]&=\E\left[ e^{i\xi(X_1+...+X_n)}\right]=\E\left[e^{i\xi X_1}\dotsm e^{i\xi X_n}\right]=\E\left[e^{i\xi X_1}\right]\dotsm\E\left[e^{i\xi X_n}\right]\\ | |||
&=\exp(-\lambda_1(1-e^{i\xi}))\dotsm \exp(-\lambda_n(1-e^{i\xi}))\\ | |||
&=\exp(-(\lambda_1+...+\lambda_n)(1-e^{i\xi}))\\ | |||
&=\exp(-Y(1-e^{i\xi})), | |||
\end{align*} | |||
</math> | |||
with <math>Y=\lambda_1+...+\lambda_n</math>. Since the characteristic function uniquely characterizes the probability distributions, we can conclude that | |||
<math display="block"> | |||
S_n\sim \Pi(Y=\lambda_1+...+\lambda_n). | |||
</math> | |||
Let now <math>X</math> be a r.v. with <math>X\sim\mathcal{N}(m,\sigma^2)</math>. Then we know | |||
<math display="block"> | |||
\E\left[e^{i\xi X}\right]=e^{i\xi m}e^{-i\sigma^2\frac{\xi^2}{2}}. | |||
</math> | |||
Now let <math>X_1,...,X_n</math> be <math>n</math> independent r.v.'s for <math>n\in \N</math>, such that <math>X_j\sim\mathcal{N}(m_j,\sigma_j^2)</math> for all <math>0\leq j\leq n</math>. Set again <math>S_n=X_1+...+X_n</math>. Therefore | |||
<math display="block"> | |||
\E\left[e^{i\xi S_n}\right]=e^{im_1\xi}e^{-\sigma^2_1\frac{\xi^2}{2}}\dotsm e^{im_n\xi}e^{-\sigma_n^2\frac{\xi^2}{2}}=e^{i(m_1+...+m_n)\xi}e^{-(\sigma_1^2+....+\sigma_n^2)\frac{\xi^2}{2}}, | |||
</math> | |||
which implies, because of the same argument as above, that | |||
<math display="block"> | |||
S_n\sim \mathcal{N}(m_1+...+m_n,\sigma_1^2+....+\sigma_n^2). | |||
</math> | |||
===Using change of variables=== | |||
Let <math>g:\R^n\to\R^n</math> be a measurable function given as <math>g(x)=(g_1(x),...,g_n(x))</math> for <math>x\in\R^n</math>. Then the jacobian of <math>g</math> is given by | |||
<math display="block"> | |||
J_g(x)=\left(\frac{\partial g_i(x)}{\partial x_j}\right)_{1\leq i,j\leq n}. | |||
</math> | |||
Recall that for <math>g:G\subset \R^n\to\R^n</math>, where <math>G</math> is a open subset of <math>\R^n</math>, with <math>J_g</math> injective such that <math>\det(J_g(x))\not=0</math> for all <math>x\in G</math>, we have for every measurable and positive map <math>f:\R^n\to\R_+</math>, or for every integrable <math>f=\one_{G}</math>, that | |||
<math display="block"> | |||
\int_{g(G)}f(y)dy=\int_Gf(g(x))\vert \det(J_g(x))\vert dx, | |||
</math> | |||
where <math>g(G)=\{y\in\R^n\mid \exists x\in G; g(x)=y\}</math>. | |||
{{proofcard|Theorem|thm-1|Let <math>(\Omega,\A,\p)</math> be a probability space. Let <math>X=(X_1,...,X_n)</math> be a r.v. on <math>\R^n</math> for <math>n\in\N</math>, having a joint density <math>f</math>. Let <math>g:\R^n\to\R^n\in C^1(\R^n)</math> be an injective measurable map, such that <math>\det(J_g(x))\not=0</math> for all <math>x\in\R^n</math>. Then <math>Y=g(X)</math> has the density | |||
<math display="block"> | |||
f_Y(y)=\begin{cases}f_X(g^{-1}(y))\vert\det{(J_{g^{-1}}(y))\vert},& y\in g(x)\\ 0,&\text{otherwise}\end{cases} | |||
</math> | |||
|Let <math>B\in\B(\R^n)</math> be a Borel set and <math>A=g^{-1}(B)</math>. Then we have | |||
<math display="block"> | |||
\p[X\in A]=\int_Af_X(x)dx=\int_{g^{-1}(B)}f_X(x)dx=\int_Bf_X(g^{-1}(x))\vert \det(J_{g^{-1}}(x))\vert dx. | |||
</math> | |||
But we know <math>\p[Y\in B]=\p[X\in A]</math>, for all <math>B\in\B(\R^n)</math> with | |||
<math display="block"> | |||
\p[Y\in B]=\int_Bf_X(g^{-1}(x))\vert\det(J_{g^{-1}}(x))\vert dx. | |||
</math> | |||
It follows that <math>Y</math> has density given by | |||
<math display="block"> | |||
f_Y(x)=f_X(g^{-1}(x))\vert \det(J_{g^{-1}}(x))\vert. | |||
</math>}} | |||
'''Example''' | |||
We got the following examples: | |||
<ul style{{=}}"list-style-type:lower-roman"><li>Let <math>X</math> and <math>Y</math> be two independent <math>\mathcal{N}(0,1)</math> distributed r.v.'s. We want to know what is the joint union distribution of <math>(U,V)=(X+Y,X-Y)</math>. Therefore, let <math>g:\R^2\to\R^2</math> be given by <math>(x,y)\mapsto (x+y,x-y)</math>. The inverse <math>g^{-1}:\R^2\to\R^2</math> is then given by <math>(u,v)\mapsto \left(\frac{u+v}{2},\frac{u-v}{2}\right)</math>. We have the following jacobian | |||
<math display="block"> | |||
J_{g^{-1}}=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&-\frac{1}{2}\end{pmatrix},\det(J_{g^{-1}})=-\frac{1}{2}. | |||
</math> | |||
Moreover we get | |||
<math display="block"> | |||
\begin{align*} | |||
f_{(U,V)}(u,v)&=f_{(X,Y)}\left(\frac{u+v}{2},\frac{u-v}{2}\right)\det(J_{g^{-1}})=\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{u+v}{2}\right)^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{u-v}{2}\right)^2}\right)\left(\frac{1}{2}\right)\\ | |||
&=\frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}}\frac{1}{\sqrt{4\pi}}e^{-\frac{v^2}{4}}. | |||
\end{align*} | |||
</math> | |||
Thus <math>U</math> and <math>V</math> are independent and <math>U\stackrel{law}{=}V\sim\mathcal{N}(0,2)</math>. | |||
</li> | |||
<li>Let <math>(X,Y)</math> be a r.v. on <math>\R^2</math> with joint density <math>f</math>. We want to find the density of <math>Z=XY</math>. In this case, consider <math>h:\R^2\to\R^2</math>, given by <math>(x,y)\mapsto xy</math>. We then define <math>g:\R^2\to\R^2</math>, given by <math>(x,y)\mapsto (xy,x)</math>. We write <math>S_0=\{(x,y)\mid x=0,y\in\R\}</math> and <math>S_1=\R^2\setminus S_0</math>. Now <math>g</math> is injective from <math>S_1</math> to <math>\R\setminus\{0\}</math> and <math>g^{-1}(u,v)=\left(v,\frac{u}{v}\right)</math>. The jacobian is thus given by | |||
<math display="block"> | |||
J_{g^{-1}}(u,v)=\begin{pmatrix}0&\frac{1}{v}\\ 1&-\frac{u}{v^2}\end{pmatrix},\det(J_{g^{-1}}(u,v))=-\frac{1}{v}. | |||
</math> | |||
Moreover we have | |||
<math display="block"> | |||
f_{(U,V)}(u,v)=f_{(X,Y)}\left(u,\frac{u}{v}\right)\frac{1}{\vert v\vert}\one_{V\not=0}. | |||
</math> | |||
Therefore, we get | |||
<math display="block"> | |||
f_U(u)=\int_{\R}f_{(X,Y)}(u,v)\frac{1}{\vert v\vert}dv. | |||
</math> | |||
</li> | |||
</ul> | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} |
Latest revision as of 01:53, 8 May 2024
The case of Sums of independent Random Variables
Let [math]X[/math] be a Poisson distributed r.v. with parameter [math]\lambda \gt 0[/math]. We already know that for all [math]\xi\in \R[/math], we then have [math]\E\left[e^{i\xi X}\right]=\exp(-\lambda(1-e^{i\xi}))[/math]. Let [math]X_1,...,X_n[/math] be [math]n[/math] independent r.v.'s for [math]n\in\N[/math] with [math]X_j[/math] being a Poisson distributed r.v. with parameter [math]\lambda_j \gt 0[/math] for all [math]1\leq j\leq n[/math]. Now let [math]S_n=\sum_{j=1}^nX_j[/math]. We want to figure out what the law of [math]S_n[/math] is. We have
with [math]Y=\lambda_1+...+\lambda_n[/math]. Since the characteristic function uniquely characterizes the probability distributions, we can conclude that
Let now [math]X[/math] be a r.v. with [math]X\sim\mathcal{N}(m,\sigma^2)[/math]. Then we know
Now let [math]X_1,...,X_n[/math] be [math]n[/math] independent r.v.'s for [math]n\in \N[/math], such that [math]X_j\sim\mathcal{N}(m_j,\sigma_j^2)[/math] for all [math]0\leq j\leq n[/math]. Set again [math]S_n=X_1+...+X_n[/math]. Therefore
which implies, because of the same argument as above, that
Using change of variables
Let [math]g:\R^n\to\R^n[/math] be a measurable function given as [math]g(x)=(g_1(x),...,g_n(x))[/math] for [math]x\in\R^n[/math]. Then the jacobian of [math]g[/math] is given by
Recall that for [math]g:G\subset \R^n\to\R^n[/math], where [math]G[/math] is a open subset of [math]\R^n[/math], with [math]J_g[/math] injective such that [math]\det(J_g(x))\not=0[/math] for all [math]x\in G[/math], we have for every measurable and positive map [math]f:\R^n\to\R_+[/math], or for every integrable [math]f=\one_{G}[/math], that
where [math]g(G)=\{y\in\R^n\mid \exists x\in G; g(x)=y\}[/math].
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math]X=(X_1,...,X_n)[/math] be a r.v. on [math]\R^n[/math] for [math]n\in\N[/math], having a joint density [math]f[/math]. Let [math]g:\R^n\to\R^n\in C^1(\R^n)[/math] be an injective measurable map, such that [math]\det(J_g(x))\not=0[/math] for all [math]x\in\R^n[/math]. Then [math]Y=g(X)[/math] has the density
Let [math]B\in\B(\R^n)[/math] be a Borel set and [math]A=g^{-1}(B)[/math]. Then we have
It follows that [math]Y[/math] has density given by
Example
We got the following examples:
- Let [math]X[/math] and [math]Y[/math] be two independent [math]\mathcal{N}(0,1)[/math] distributed r.v.'s. We want to know what is the joint union distribution of [math](U,V)=(X+Y,X-Y)[/math]. Therefore, let [math]g:\R^2\to\R^2[/math] be given by [math](x,y)\mapsto (x+y,x-y)[/math]. The inverse [math]g^{-1}:\R^2\to\R^2[/math] is then given by [math](u,v)\mapsto \left(\frac{u+v}{2},\frac{u-v}{2}\right)[/math]. We have the following jacobian
[[math]] J_{g^{-1}}=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&-\frac{1}{2}\end{pmatrix},\det(J_{g^{-1}})=-\frac{1}{2}. [[/math]]Moreover we get[[math]] \begin{align*} f_{(U,V)}(u,v)&=f_{(X,Y)}\left(\frac{u+v}{2},\frac{u-v}{2}\right)\det(J_{g^{-1}})=\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{u+v}{2}\right)^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{u-v}{2}\right)^2}\right)\left(\frac{1}{2}\right)\\ &=\frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}}\frac{1}{\sqrt{4\pi}}e^{-\frac{v^2}{4}}. \end{align*} [[/math]]Thus [math]U[/math] and [math]V[/math] are independent and [math]U\stackrel{law}{=}V\sim\mathcal{N}(0,2)[/math].
- Let [math](X,Y)[/math] be a r.v. on [math]\R^2[/math] with joint density [math]f[/math]. We want to find the density of [math]Z=XY[/math]. In this case, consider [math]h:\R^2\to\R^2[/math], given by [math](x,y)\mapsto xy[/math]. We then define [math]g:\R^2\to\R^2[/math], given by [math](x,y)\mapsto (xy,x)[/math]. We write [math]S_0=\{(x,y)\mid x=0,y\in\R\}[/math] and [math]S_1=\R^2\setminus S_0[/math]. Now [math]g[/math] is injective from [math]S_1[/math] to [math]\R\setminus\{0\}[/math] and [math]g^{-1}(u,v)=\left(v,\frac{u}{v}\right)[/math]. The jacobian is thus given by
[[math]] J_{g^{-1}}(u,v)=\begin{pmatrix}0&\frac{1}{v}\\ 1&-\frac{u}{v^2}\end{pmatrix},\det(J_{g^{-1}}(u,v))=-\frac{1}{v}. [[/math]]Moreover we have[[math]] f_{(U,V)}(u,v)=f_{(X,Y)}\left(u,\frac{u}{v}\right)\frac{1}{\vert v\vert}\one_{V\not=0}. [[/math]]Therefore, we get[[math]] f_U(u)=\int_{\R}f_{(X,Y)}(u,v)\frac{1}{\vert v\vert}dv. [[/math]]
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].