# Is not always the accumulated probability until the expected value equal to 1/2?

My problem is that doing calculations with exponential distribution, the result I get is not 1/2, and I think it should be that according to the intuitive concept of the expected value. The density function for an exponential distribution is:

f(y) = (1/β) * e^(-y/β)
0 < y <∞

The average is β (that is known in advance). Calculating the accumulative probability to the average:

F(β) = ∫{0 < y < β} (1/β)*e^(-y/β) dy

F(β) = - ∫{0 < y < β} e^(-y/β) * (-1/β) dy

F(β) = - [e^(-y/β)] {y from 0 to β}

F(β) = - [e^(-β/β) - e^(0/β)]

F(β) = 1 - e^(-1)

F(β) = 1 - 1/e

F(β) = (e-1)/ e

I think the error is conceptual-

• Consider a raffle with one million \$1 tickets and a one million dollar prize for the winner. The expected value of a ticket after entering the raffle is \$1, but the probability of making less than the expected value is $$1 - \frac{1}{1000000} > \frac{1}{2}.$$ Oct 27, 2015 at 3:16
• @EricTressler: I use the exploited $1000$ workers at a factory, and the owner who makes a billion from their pain. Oct 27, 2015 at 3:24

You are asking whether the median of a random variable $X$, that is, the place $m$ such that $\Pr(X\le m)=\frac{1}{2}$, is the same as the mean $E(X)$. For a symmetric distribution, like the normal, the median is equal to the mean (if the mean exists). But for "most" distributions, including the exponential, the median is not equal to the mean.
The "accumulated probability" until $E(X)$ is $\frac{1}{2}$ precisely if the median is equal to the mean.
• I still do not see it. If we make the graph, the area under the curve in total should be 1, and what I thought is, regardless of whether it is symmetric, if we divide it by the value in $x = E(x)$, on both sides left area should be equal to 1/2. Oct 27, 2015 at 3:31
• In your case $\beta$ is the mean. You were calculating $Pr(Y\le \beta)$, got a result other than $1/2$, and were surprised. That's because you were expecting that the probability would be $1/2$, that is, you were expecting that $\beta$ would be the median. It isn't. Oct 27, 2015 at 3:38
• Now I will calculate the median $m$. We want $\Pr(Y\le m)=1/2$. After a little manipulation we get $e^{-m/\beta}=1/2$. Taking the $\ln$ of both sides we get $m=(\ln 2)\beta$. So that's how mean and median are related for the exponential. The median is quite a bit less than the mean. Oct 27, 2015 at 3:42