Let $X$ be a random variable with finite variance. I am trying to show if $P(|X-E(X)|\leq x)=1$ then $V(X)\leq x^2$. Could somebody please help me correct my working?
$|X-E(X)|\leq x\iff(X-E(X))^2\leq x^2\iff E[(X-E(X))^2]\leq E[x^2]$ $\iff V(X)\leq E[x^2]$ by monotonicity of expectation.
So $P(|X-E(X)|\leq x)=P((X-E(X))^2\leq x^2)=P(E[(X-E(X))^2]\leq E[x^2])=P(V(X)\leq E[x^2])=1$
Then I am stuck. How can we get rid of the probability and arrive at $V(X)\leq x^2$?
Thanks.
Let $Y=X-\mathrm{E}[X]$. Then $$ V(X)=\mathrm{E}[Y^2]=\mathrm{E}[Y^2\mathbf{1}_{|Y|\leqslant x}]\leqslant \mathrm{E}[x^2\mathbf{1}_{|Y|\leqslant x}]=x^2P(|Y|\leqslant x)=x^2. $$