Q: Proving for $E(X^2)>1$ , then $P(X=0)<1$ If $X$ is a random variable with $E(X^2)>1$ then $P(X=0)<1$ What I have done so far is I tried to use Markov's Inequality: $$P(X=0) \le \frac{E(X^2)}{E(X^2)} \le 1$$ But I'm not sure if this is the right way of proving it, because I'm not sure if this makes sense to say $E(X^2) > 1$ which means it's not a constant. Any help would be greatly appreciated, thanks A: If $E(X^2)>1$ then $E(X)=0$. The rest is routine verification. A: $E(X)=0$ is not really possible, since $E(X)=0$ only for constant random variable. A: If $E[X^2]>1$, then $E[X^2]=1+\sigma^2>1$ where $\sigma^2>0$, so $E(X^2-1)=E(X^2)>0$ then $P(X=0)<1$. This can be shown more rigorously using the fact that if $E[X]=0$, then $X$ is a constant random variable, so $E[X^2]=\sigma^2$ and $E[X^2-1]=\sigma^2-1$ is positive, so $P(X=0)<1$ A: $$ E(X^2)>1\iff E(X^2)-1>0\iff\sigma^2>1 $$ so $\sigma^2$ is positive. Then by Markov's inequality $$ P(X=0)=\mathbb{P}(X^2-1\le0)\le\frac{E(X^2-1)}{E(X^2)-1}=\frac{\sigma^2-1}{\sigma^2}=1-\frac{1}{\sigma^2} $$ with $\sigma^2>1$. Remark: As mentioned by others, $E(X)=0$ is impossible. However, $\sigma^2>1$ is possible, and in fact typical. This can be seen from $$ E(X^2)=E(X)^2+2\sigma^2 $$ where $E(X)^2$ is the expectation of $X^2$ if $X$ were constant. Thus if $E(X)^2$ is small, $\sigma^2$ will be large. For example if $X$ is uniformly distributed on $[0,1]$, then $\sigma^2=\frac{1}{12}$. If $X$ is a Cauchy distributed random variable, then $\sigma^2=+\infty$ and again $P(X=0)=1$. If $X$ is a normal distributed random variable, then $P(X=0)=0$. If $X$ is exponentially distributed with mean $\mu$, then $$ E(X^2)=\frac{2}{\mu^2+2\mu} $$ so $\sigma^2=\frac{1}{\mu^2+2\mu}$. These examples can be generalized to any distribution with a positive variance and non-negligible density at the lower boundary of its support. As I hinted in the comments, $\sigma^2$ must be positive for such a distribution. This is again proven by contradiction. If $\sigma^2=0$, then $X^2=X$, and $X$ is constant, hence $E(X)=0$, contradicting our assumption.