Let $|X| ≤ b $ be a random variable. Show $\forall a \in [0,b)$ $ \mathbb{P}(|X|>a)≥\frac{\mathbb{E}[X]-a}{b-a}$

Question

I've been set the following question, but I think the question is wrong.

Let $|X| ≤ b $ be a random variable. Show $\forall a \in [0,b)$ $ \mathbb{P}(|X|>a)≥\frac{\mathbb{E}[X]-a}{b-a}$

Working: Let $a \in [0,b) $ be arbitrary.

By the total law of probability for expectation $$\mathbb{E}[|X|] = \mathbb{E}[|X|:|X|>a]\mathbb{P}(|X|>a) + \mathbb{E}[|X|:|X|≤a]\mathbb{P}(|X|≤a)$$ Noting that $\mathbb{P}(|X|≤a) = 1-\mathbb{P}(|X|>a)$, we have $$\mathbb{E}[|X|] = (\mathbb{E}[|X|:|X|>a]-\mathbb{E}[|X|:|X|≤a])\mathbb{P}(|X|>a) + \mathbb{E}[|X|:|X|≤a]$$ and so $$\mathbb{P}(|X|>a) = \frac{\mathbb{E}[|X|]-\mathbb{E}[|X|:|X|≤a]}{\mathbb{E}[|X|:|X|>a]-\mathbb{E}[|X|:|X|≤a]}$$

Now $\mathbb{P}(|X|>a)≥\min \text{RHS}$, and noting that $0≤\mathbb{E}[|X|:|X|≤a]≤a$ and $a<\mathbb{E}[|X|:|X|>a]≤b$ should yield the answer.

However, to minimise our fraction, the denominator must be large and so the maximum value of the denominator is $b$, not the $b-a$ the question asks for.

Is it possible that the question setter has mistakenly thought $w<x, y<z \implies w-y<x-z$, or have I missed something here!?

Answer 1

The inequality is correct. As a start, convince yourself (by considering cases if necessary) that $X-a\le(b-a)1_{\{|X|>a\}}$.

Answer 2

The claim is equivalent to

$$\mathbb{E}(X) \leq a + (b-a)\mathbb{P}(|X| >a).$$

To prove this note that

$$\begin{align*} |\mathbb{E}(X)| &\leq \mathbb{E}(|X|) = \mathbb{E}(|X| 1_{\{|X| \leq a\}}) + \mathbb{E}(|X| 1_{\{a <|X| \leq b\}}) \\ &\leq a \mathbb{P}(|X| \leq a) + b \mathbb{P}(a < |X| \leq b). \end{align*}$$

Since

$$\mathbb{P}(|X| \leq a) =1-\mathbb{P}(a<|X| \leq b)$$

and $\mathbb{P}(a < |X| \leq b) = \mathbb{P}(a<|X|)$, we find

$$\mathbb{E}(X) \leq a + (b-a)\mathbb{P}(|X| >a).$$