Here is the question: Suppose $X$ is a $\Bbb R$-valued random variable. Show that for all $\varepsilon > 0$, there exists a bounded random variable $Y$ such that $P (X \neq Y ) <\varepsilon$.
For every $\Bbb R$-valued random variable $X$ and $\epsilon > 0$, $P (X \neq Y ) <\epsilon$ for some bounded random variable $Y$
2
$\begingroup$
probability-theory
random-variables
1 Answers
1
Try $Y=X\,\mathbf 1_{|X|\leqslant x}$ for $x$ large enough. Then $\mathbb P(Y\ne X)=\mathbb P(|X|\gt x)$. Furthermore, $\mathbb P(|X|\gt x)\to0$ when $x\to+\infty$, hence for $x$ large enough, $Y$ solves the question.
-
0I presume that $P(X=Y)$ in the question is a typo and should read $P(X\ne Y)$. – 2012-10-12
-
0What about the part {X>x}. Could you explain a little bit more? – 2012-10-13
-
0Sorry but I do not understand the question in your comment. If *explain a little bit more* means *provide a full solution one can hand as it is to one's teacher*, the answer is *no*. Otherwise, please explain the part you do not understand. – 2012-10-13
-
0It's obviously Y can cover $X\,1_{\left|X\right|\leq x}$ for arbitrary x. But how do you deal with the part $X\,1_{\left|X\right|>x}$? How can it be less than any positive number when you fix a x? Thanks! – 2012-10-13
-
0*Cover*? *The part*? *Fix* $x$? See the expanded version of my answer. – 2012-10-13
-
0That's correct. I misunderstood the question. I thought I should find a Y for all all ε>0, P(X≠Y)<ε. – 2012-10-16
-
0@Did would you explain what do you mean by stating $1_{|X| \le x}$ – 2017-08-15
-
0@PhilJones https://en.wikipedia.org/wiki/Indicator_function – 2017-08-15
-
0@Did would you give a brief explanation on how Y = $X 1_{|X| \le x}$ is a bounded random variable... I'm having a bit of difficulty in understanding so... – 2017-08-15
-
0@PhilJones For every $\omega$, $|Y(\omega)|\leqslant x$. – 2017-08-15
-
0@Did Yeppp, now understood. – 2017-08-15