Can we say that the value of the cumulative distribution function at the mean F(X< Mean) is always 0.5 for all kind of distributions (even ones that are not symmetric)?
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4 Answers
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No, this is false. That point is the median, and it is not equal to the mean in all cases, for example, an exponential distribution.
$$ X\sim \exp(1)\\ \mathbb E[X]=1\\ \operatorname{median}(X)=\log 2\approx 0.69 $$
We can simulate this in software, such as R.
set.seed(2021)
X <- rexp(10000, 1)
mean(X) # approximately 1
median(X) # approximately 0.69
EDIT
The asymmetric Laplace distribution is an example where the distribution extends over the entire real line, not just the positive numbers. (In more technical language, the PDF has support on all of $\mathbb R$.)
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2$\begingroup$ Funny we came up with the same distribution for a counterexample! $\endgroup$– B.LiuCommented Jan 19, 2022 at 10:28
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The mean is the point where the area left and right between the CDF and the vertical line through the mean are equal.
For symmetric distributions, the left and the right side/area have the same shape and you get that the mean is at the point where the CDF is equal to 0.5 (also the median).
For non-symmetric distributions the shapes are different, so this equality of area does not need to be true (obviously it can be equal, but it does not need to be equal).
For any distribution where the mean and median is different, you have that the CDF of the mean is not 0.5.
An example is a non-central t-distribution which is plotted below.
answered Jan 19, 2022 at 11:13
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1$\begingroup$ So, if $\mu = E[X]$ and $F$ is continuous, we have $\int_{-\infty}^\mu F(t)dt = \int_{\mu}^\infty (1-F(t))dt$?. Do you have a quick reference? I do not remember this result. $\endgroup$ Commented Jan 19, 2022 at 12:01
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2$\begingroup$ I do not know a direct statement of $\int_{-\infty}^\mu F(t)dt = \int_{\mu}^\infty (1-F(t))dt$ I view it intuitively as integrating the quantile function $E[X] = \int_0^1 Q_X(p) dp$ or this expression $\endgroup$ Commented Jan 19, 2022 at 12:06
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3$\begingroup$ It follows from$$\int_{-\infty}^\mu F(t)\text dt=\mu F(\mu)-\int_{-\infty}^\mu t \text dF(t)$$and$$\int^{\infty}_\mu (1-F(t))\text dt=-\mu +\mu F(\mu)+\int^{\infty}_\mu t \text dF(t)=\mu F(\mu)-\int_{-\infty}^\mu t \text dF(t)$$ $\endgroup$– Xi'anCommented Jan 19, 2022 at 12:08
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1$\begingroup$ @Xi'an very clean, thank you! $\endgroup$ Commented Jan 19, 2022 at 12:27
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1$\begingroup$ @Lucas We have $$E[X] = \int_0^1 Q_X(p) dp$$ and we can change this like the following which is considering the integral of the quantile function with a shift, which is like the area between $F(x)$ and the vertical line through $E[X]$ $$0 = \int_0^1 Q_X(p)-E[X] dp$$ or $$ \underbrace{\int_0^m Q_X(p)-E[X] dp}_{\text{negative of area leftside}} + \underbrace{\int_m^1 Q_X(p)-E[X] dp}_{area rightside}$$ where $m$ is the quantile of the expectation value. $\endgroup$ Commented Jan 19, 2022 at 12:29
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As others have said, no. In fact, it can be any number in $(0, 1$).
Let $\epsilon \, \in \, (0,1)$, consider the random variable $X$ such that $\mathbb{P}(X = -1) = \epsilon$ and $\mathbb{P}(X = \frac{\epsilon}{1-\epsilon}) = 1 - \epsilon$. The mean is
$$E[X] = -\epsilon + (1-\epsilon)\frac{\epsilon}{1-\epsilon} = 0 \quad, $$
and $F(0) = \epsilon$.
As observed in the comments, $1$ can also be attained by the CDF if we consider a constant random variable.
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2$\begingroup$ The point 1 is also included if you consider degenerate distributions (eg. constants). $\endgroup$ Commented Jan 19, 2022 at 11:17
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No.
A counterexample will be $X \sim Exp(1)$. The mean of $X$ is 1, and $P(X < 1)$ is $1 - 1/e$, around 0.63.
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$\begingroup$ This hurts my brain. I get that the math checks out but it's not very intuitive. $\endgroup$ Commented Apr 27, 2022 at 15:42