0
$\begingroup$

The following question is taken from here problem $2.5:$

Exercise $2.5:$ Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $f(x) \geq 0$ for all $x$ and $$\int_{-\infty}^\infty f(x) \, dx = 1.$$ For $r\geq0,$ let $$I_n(x) = \int\dots\int_{x_1^2+x_2^2+\dots+x_n^2 \leq r^2} f(x_1) f(x_2)\dots f(x_n) \, dx_1 \, dx_2 \dots \, dx_n.$$ Find $\lim_{n\to\infty}I_n(x)$ for a fixed $r.$

The answer given is $0.$ I think I need to use $n$-dimensional spherical substitution to reduce the problem over an $n$-dimension sphere.

However, I have no idea how to use it. Any hint would be appreciated.

$\endgroup$
  • $\begingroup$ first step: have you looked up the formulas for $n$-dimensional spherical substitution and looked at the examples? $\endgroup$ – Brevan Ellefsen Aug 25 '17 at 15:43
  • 1
    $\begingroup$ A short answer is that $$I_n(r)=P(X_1^2+\cdots+X_n^2\leqslant r^2)$$ where $(X_k)$ is i.i.d. with PDF $f$ hence, say by the weak law of large numbers, $$I_n(r)\to0$$ The continuity of $f$ is not needed. $\endgroup$ – Did Aug 25 '17 at 16:40
0
$\begingroup$

$f$ satisfies definition for probability density function. So the problem can be reformulated as given $X_i \overset{iid}{\sim} X$, \begin{align*} I_n &= E[I(X_1^2+\dotsb, X_n^2 \le r^2) ]\\ &= P(X_1^2+\dotsb, X_n^2 \le r^2)\\ &= P(nX^2 \le r^2)\\ &= P(X^2 \le \frac{r^2}{n})\\ \lim_{n\to\infty} I_n &= P(X^2 \le 0)\\ &= 0 \end{align*} The last few lines are a bit hand wavy, but I hope this helps.

$\endgroup$
  • 1
    $\begingroup$ The identity $$P(X_1^2+\dotsb, X_n^2 \le r^2)= P(nX^2 \le r^2)$$ is of course quite wrong. $\endgroup$ – Did Aug 25 '17 at 16:42
-1
$\begingroup$

Hint: More generally, suppose $f$ is continuous and nonnegative on $[-1,1],$ with $f\le M$ there. Then $I_n$ is bounded above by $M^n$ times the $n$-dimensional volume of $B(0,r).$ That volume is $r^n$ times the $n$-dimensional volume of $B(0,1).$ So everything depends on finding a good estimate of $V_n(B(0,1)).$

$\endgroup$
  • $\begingroup$ Why the downvote? $\endgroup$ – Idonknow Aug 26 '17 at 0:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.