$P(X \geq t) \leq \frac{E(X^2)}{E(X^2) + t^2}$, where $E(X) = 0$ and $E(X^2)$ is finite.

Question

I'm trying to prove the above inequality. I've tried the following; note I make use of the Cauchy-Schwarz Inequality and the fact that $I$ is an indicator function:

$$|(t-X)I(t-X>0)|^2 \leq \\E[(t-X)^2]E[(I(t-X>0))^2] \\ = (t^2 + E(X^2))P(t-X >0) =\\ t^2 + E(X^2),$$ since $(t-X) \leq (t-X)I(t-X>0) \implies P(t-X>0) \geq 1 \implies P(t-X>0)=1.$

Now I'm a bit stuck. I have tried relating $t^2 + E(X^2)$ to $|t-X|^2$ but am not making progress. I'm trying to think of how we can bring $P(X\geq t)$ into the mix as well, since eventually we will obviously need it!

EDIT: $t \geq 0$.

Answer 1

If $t$ is not necessarily positive, then I think this is wrong.

Let $X\equiv 0$.
For $t=-1$ you get $$1=\mathbb P(X\geq -1)\leq 0 = \frac{0}{0+1}\implies 1\leq 0\implies \unicode{x21af}$$

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Another example would be to take $X\sim Ber(1/2)$, and pick $t= -1$.
You get: $$1=\mathbb P(X\geq -1)\leq \frac{1}{2} = \frac{1}{1+1}\implies 1\leq\frac{1}{2}\implies \unicode{x21af}$$

I think there are a lot more cases, where this does not apply.

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Answer 2

$ \begin{aligned} &(t-X)\leq (t-X)\textbf{1}\{t-X>0\}\\ \Rightarrow&0\leq t=\mathbb{E}[t-X]\leq \mathbb{E}[(t-X)\textbf{1}\{t-X>0\}]\\ \Rightarrow& (\mathbb{E}[t-X])^2\leq (\mathbb{E}[(t-X)\textbf{1}\{t-X>0\}])^2 \end{aligned} $

$ \begin{aligned} t^2&=(\mathbb{E}[t-X])^2\\ &\leq (\mathbb{E}[(t-X)\textbf{1}\{t-X>0\}])^2\\ &\leq \mathbb{E}[(t-X)^2]\mathbb{E}[\textbf{1}\{t-X>0\}^2]\\ &=\mathbb{E}[(t-X)^2]\mathbb{E}[\textbf{1}\{t-X>0\}]\\ &\leq \mathbb{E}[(t-X)^2]\cdot \mathbb{P}(t-X>0)\\ &=(t^2+\mathbb{E}[X^2])\cdot \mathbb{P}(X<t) \end{aligned} $

We can reformulate this to $\mathbb{P}(X\geq t)\leq \frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]+t^2}.$