A Target-Based Foundation for the “Hard-Easy Effect” Bias

Abstract

The “hard-easy effect” is a well-known cognitive bias on self-confidence calibration that refers to a tendency to overestimate the probability of success in hard-perceived tasks, and to underestimate it in easy-perceived tasks. This paper provides a target-based foundation for this effect, and predicts its occurrence in the expected utility framework when utility functions are S-shaped and asymmetrically tailed. First, we introduce a definition of hard-perceived and easy-perceived task based on the mismatch between an uncertain target to meet and a suitably symmetric reference point. Second, switching from a target-based language to a utility-based language, we show how this maps to equivalence between the hard-perceived target/gain seeking and the easy-perceived target/loss aversion. Third, we characterize the agent’s miscalibration in self-confidence. Sufficient conditions for acting according to the “hard-easy effect” and the “reversed hard-easy effect” biases are set out. Finally, we derive sufficient conditions for the “hard-easy effect” and the “reversed hard-easy effect” to hold. As a by-product we identify situations in enterprise risk management where misconfidence in judgments emerges. Recognizing these cognitive biases, and being mindful of to be normatively influenced by them, gives the managers a better framework for decision making.

Appendix

1.1 Proof of Proposition 1

We will prove that if T is a hard-perceived/easy-perceived task with respect to m its opposite-perceived variable Y is an easy-perceived/hard-perceived task.

Let $\overline{\mathit{\mathsf{T}}}$ a hard-perceived target with c.d.f. $\overline{\mathit{\mathsf{u}}}$. Construct the variable Y with c.d.f. $\underline {\mathit{\mathsf{u}}}$ such that $\mathit{\mathsf{Y}}-\mathit{\mathsf{m}}=-\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)$. It follows that

$$ \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)=\mathsf{1}-\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)\kern0.75em \mathsf{f}\mathsf{o}\mathsf{r}\ \mathsf{all}\kern0.5em \mathit{\mathsf{x}}\ge \mathsf{0}. $$

(3)

In fact

$$ \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)=\mathit{\mathsf{P}}\left(\mathit{\mathsf{Y}}\le \mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)=\mathit{\mathsf{P}}\left(\mathit{\mathsf{Y}}-\mathit{\mathsf{m}}\le \mathit{\mathsf{x}}\right)=\mathit{\mathsf{P}}\left(-\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)\le \mathit{\mathsf{x}}\right)=\mathit{\mathsf{P}}\left(\overline{\mathit{\mathsf{T}}}\ge \mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)=\mathsf{1}-\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right) $$

And analogously

$$ \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)=\mathsf{1}-\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)\kern0.5em \mathsf{f}\mathsf{o}\mathsf{r}\ \mathsf{all}\kern0.5em \mathit{\mathsf{x}}\ge \mathsf{0}. $$

(4)

Summing up (3) and (4), we get

$$ \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)+\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)=\mathsf{1}-\left[\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)+\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)-\mathsf{1}\right] $$

Since $\overline{T}$ is a hard-perceived target, then $\left[\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)+\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)-\mathsf{1}\right]\ge \mathsf{0}$, so

$\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)+\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)\le \mathsf{1}$ for all $\mathit{\mathsf{x}}\ge \mathsf{0}$,

Then by Definition 1(b) Y is an easy-perceived target.

The two opposite-perceived targets have the same median m. In fact, if $\mathit{\mathsf{x}}=\mathsf{0}$, it follows $\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}\right)=\mathsf{1}-\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}\right)=\mathsf{0.5}$. If T is an easy-perceived target, the relations are reversed and the opposite-perceived variable Y can be proved to be hard-perceived.

1.2 Proof of Theorem 3

Let $\overline{\mathit{\mathsf{T}}}$ be a hard-perceived target with c.d.f. $\overline{\mathit{\mathsf{u}}}$ and $\underline {\mathit{\mathsf{T}}}$ its opposite-perceived variable with c.d.f. $\underline {\mathit{\mathsf{u}}}$, such that $\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}=-\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)$. We will prove that the relation between $\overline{\mathit{\mathsf{u}}}$ and $\underline {\mathit{\mathsf{u}}}$ switches at m. Specifically, $\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)\le \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)$ and $\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)\le \overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)$ for any $\mathit{\mathsf{x}}\ge \mathsf{0}$.

Abdous and Theodorescu (1998, Eq. (2), p. 357) set the equivalence between the van Zwet conditions (1979, (1.2), p. 1) and the first stochastic order between the two tails of T around m, it holds

$\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)+\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)<\mathsf{1}$ is equivalent to ${\left(\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{+}{\succ}_{\mathit{\mathsf{s}}\mathit{\mathsf{t}}}{\left(\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{-}$ and

$\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)+\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)\ge \mathsf{1}$ is equivalent to ${\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{-}{\succ}_{\mathit{\mathsf{s}}\mathit{\mathsf{t}}}{\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{+}$,

where ${\mathit{\mathsf{X}}}^{\pm }$ denote the positive (negative) part of X.

Since $\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}=-\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)$ we have ${\left(\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{+}={\left(-\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)\right)}^{+}={\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{-}$, then

${\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{-}{\succ}_{\mathit{\mathsf{s}}\mathit{\mathsf{t}}}{\left(\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{-}$, so $\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)\le \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)$ for $\mathit{\mathsf{x}}\ge \mathsf{0}$

And since ${\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{-}={\left(-\left(\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)\right)}^{-}={\left(\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{+}$, we have

${\left(\underline {\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{+}{\succ}_{\mathit{\mathsf{s}}\mathit{\mathsf{t}}}{\left(\overline{\mathit{\mathsf{T}}}-\mathit{\mathsf{m}}\right)}^{+}$, so $\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)\le \overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)$ for $\mathit{\mathsf{x}}\ge \mathsf{0}$

The above can be rewritten as $\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)\le \overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)$ for $s\le m$ and $\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)\ge \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)$ for $\mathit{\mathsf{s}}\ge \mathit{\mathsf{m}}$.

Let ${\mathit{\mathsf{u}}}_{\mathsf{0}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)=\frac{\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)+\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)}{\mathsf{2}}$ for any x. By construction, it holds

$\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)\le {\mathit{\mathsf{u}}}_{\mathsf{0}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)\le \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}-\mathit{\mathsf{x}}\right)$ and $\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)\ge {\mathit{\mathsf{u}}}_{\mathsf{0}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)\ge \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{m}}+\mathit{\mathsf{x}}\right)$ for $\mathit{\mathsf{x}}\ge \mathsf{0}$.

Let now quantify the probability that the lottery X outperforms the target T. Consider the lottery X such that:

(a) The outcomes of X belong on $\left[\mathit{\mathsf{m}},+\infty \right)$ . Since $\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)={\mathit{\mathsf{u}}}_{\mathsf{0}}\left(\mathit{\mathsf{s}}\right)\le \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)=\mathsf{0}$ for all $\mathit{\mathsf{s}}<\mathit{\mathsf{m}}$, following relation holds

$\mathsf{E}\left(\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{X}}\right)\right) < \mathsf{E}\left({\mathit{\mathsf{u}}}_{\mathsf{0}}\left(\mathit{\mathsf{X}}\right)\right)<\mathsf{E}\left(\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{X}}\right)\right)$.

Then

$\mathit{\mathsf{P}}\left(\mathit{\mathsf{X}}\ge \underline {\mathit{\mathsf{T}}}\right)<\mathit{\mathsf{P}}\left(\mathit{\mathsf{X}}\ge {\mathit{\mathsf{T}}}_{\mathsf{0}}\right)<\mathit{\mathsf{P}}\left(\mathit{\mathsf{X}}\ge \overline{\mathit{\mathsf{T}}}\right)$.

So if the lottery X promises high stakes, all above or equal to the external reference point m, then the agent is prone to under confidence in easy-perceived tasks, and to overconfidence in hard-perceived tasks. That risk attitude follows the “hard-easy effect.”

(b) The outcomes of X belong on $\left(-\infty, \mathit{\mathsf{m}}\right]$ . Since $\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)={\mathit{\mathsf{u}}}_{\mathsf{0}}\left(\mathit{\mathsf{s}}\right)\le \underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{s}}\right)=\mathsf{0}$ for all $\mathit{\mathsf{s}}>\mathit{\mathsf{m}}$, following relation holds

$\mathsf{E}\left(\overline{\mathit{\mathsf{u}}}\left(\mathit{\mathsf{X}}\right)\right) < \mathsf{E}\left({\mathit{\mathsf{u}}}_{\mathsf{0}}\left(\mathit{\mathsf{X}}\right)\right)<\mathsf{E}\left(\underline {\mathit{\mathsf{u}}}\left(\mathit{\mathsf{X}}\right)\right)$.

Then

$\mathit{\mathsf{P}}\left(\mathit{\mathsf{X}}\ge \overline{\mathit{\mathsf{T}}}\right)<\mathit{\mathsf{P}}\left(\mathit{\mathsf{X}}\ge {\mathit{\mathsf{T}}}_{\mathsf{0}}\right)<\mathit{\mathsf{P}}\left(\mathit{\mathsf{X}}\ge \underline {\mathit{\mathsf{T}}}\right)$.

So if the lottery X promises bad outcomes, all below or equal to the external reference point m, then the agent is prone to under confidence in hard-perceived tasks, and to overconfidence in easy-perceived tasks. That risk attitude follows the “reversed hard-easy effect”.