Let $X_1$ and $X_2$ be i.i.d. random variable.

For $\alpha\in(0,1)$, we define a new RV $X$ in a way that $X=\alpha X_1+(1-\alpha)X_2$.

For some $z$, I am interested in the comparison of the two values

$$E[X_1|X_1\geq z]~\textrm{and}~E[X_1|X\geq z].$$

The former value measures the exact expected value given that $X_1\geq z$, while the latter measures the expected value when only noisy information about $X_1$ is available. (namely $X_1\geq \frac{z-(1-\alpha)X_2}{\alpha}$).

For $X_1$ and $X_2$ uniform RV over [0,1], I found that we can have both cases of $E[X_1|X_1\geq z]<E[X_1|X\geq z]$ and $E[X_1|X_1\geq z]>E[X_1|X\geq z]$ depending on the level of $z$.

Especially, if $z$ is relatively large, we have $E[X_1|X_1\geq z]<E[X_1|X\geq z]$. On the other hand, if $z$ is low, we have the latter inequality.

Initially, I expected only "$E[X_1|X_1\geq z]>E[X_1|X\geq z]$" should be true regardless of the level of $z$ because $X\geq z$ is less informative than $X_1\geq z$.

Do you have any intuitive explanation why the both cases can happen? and Why we have the former inequality for higher $z$'s?

  • $\begingroup$ Can you give your counterexample? For $X1$ and $X2$ independent uniform variables on the unit interval and $\alpha=.5$, I get that the first is $(1+z)/2$ and the second is $(2z+1)/3$, so the former is always larger as intuition expects for $z<1$ (which z has to be in order for the possible set to be nonempty). $\endgroup$
    – Eric
    Aug 20, 2022 at 17:21
  • $\begingroup$ @Eric For the same independent uniform and for $\alpha=0.5$, I also got the former larger than the latter. But for $\alpha=0.9$ and $z=0.8$, we have $E[X_1|X_1\geq0.8]=0.9$ and $E[X_1|X\geq 0.8]=0.91358$. For the range of $1-\alpha<z<\alpha$, we have $E[X_1|X\geq z]=\frac{2\alpha^2-3\alpha z+2\alpha-3z^2+3z-1}{3\alpha(\alpha-2z+1)}$ which can be larger than $(1+z)/2$ $\endgroup$
    – Andeanlll
    Aug 21, 2022 at 0:07

2 Answers 2


To help understanding, I will set $\alpha = 0.1$ and $z = 4.5$

Let's think of an ordinary dice($1$ to $6$).

$X_1, X_2$ are numbers of $i$th rolling.

First, let's compute $E[X_1\vert X_1 > z]$.

$E[X_1\vert X_1 > z] = E[X_1\vert X_1 > 4.5] = E[X_1\vert X_1 \in \{5, 6\}] = \frac{5+6}{2} = 5.5$

Second, let's think of $\alpha X_1 + (1-\alpha)X_2$.

$\alpha X_1 + (1-\alpha)X_2 > z $ becomes $0.1X_1 + 0.9X_2 > 4.5$

which is same to $X_1 + 9X_2 > 45$

Then, $X_1$ can be any number from $1$ to $6$ if $X_2 \geq 5$.

Thus, $E[X_1\vert \alpha X_1 + (1-\alpha)X_2 > z]$ would be less than $5.5$.

(Although it's just guesswork not from the precise calculation, we can accept that it is indeed less than $5.5$ cuz $X_1$ varies from $1$ to $6$)

So, what I want to say from this example is that

$X_1>z$ limits the domain of $X_1$ that it must be larger than $z$,

but $\alpha X_1 + (1-\alpha)X_2 > z $ doesn't limit or limit the domain of $X_1$ partly for some $\alpha$.

(e.g. $X_2$ helped $X_1$ to have small numbers by having larger number($5, 6$) itself)

This difference makes the results.

++ If you set $\alpha = 0.1$ and $z$ properly, so that $\alpha X_1 + (1-\alpha)X_2 > z $ implies $X_1 = 6$, you would get the opposite case.


Not sure if this is an intuitive explanation, but firstly '$X_1\ge z$' does not contain more information than '$X\ge z$'. I'm not sure exactly what you mean by containing more information but for example, $\{X\ge z\}$ is not a subset of $\{X_1\ge z\}$. If $X\ge z$ then either $X_1\ge z$ or $X_2\ge z$ but we don't know which one. In addition, $X$ is not a function of $X_1$ nor is $X_1$ a function of $X$ so we do not have inclusions of $\sigma(X)$ into $\sigma(X_1)$ or vice versa. Based on this, one cannot really compare the conditional expectation values.

The other thing is just because the conditioning event gets larger, the conditional expectation need not get larger. A conditional expectation gives us the average value over the region. Expanding into certain regions may decrease your conditional expectation value.


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