# Thread: expected value of positive discrete RV

1. ## expected value of positive discrete RV

I'm required to show that if $\displaystyle X\geq 0$ then $\displaystyle \sum_{n=1}^{\infty}\mathbb{P}(X\geq n) \leq \mathbb{E}[X] \leq 1 + \sum_{n=1}^{\infty}\mathbb{P}(X\geq n)$

For the continuous case it was easy enough to express $\displaystyle \mathbb{E}[X]$ in terms of the CDF and the result followed from there...

But for the discrete case I'm a little stuck as to how to start...

$\displaystyle \mathbb{E}[X] = \sum_{i=0}^{m}x_iP(X=x_i)$
$\displaystyle \sum_{n=1}^\infty\mathbb{P}(X\geq n) = \sum_{n=1}^\infty\sum_{i=n}^m\mathbb{P}(X=x_i)$

hmmmm....

ay help would be much appreciated

2. Hello,
Originally Posted by RanDom
I'm required to show that if $\displaystyle X\geq 0$ then $\displaystyle \sum_{n=1}^{\infty}\mathbb{P}(X\geq n) \leq \mathbb{E}[X] \leq 1 + \sum_{n=1}^{\infty}\mathbb{P}(X\geq n)$

For the continuous case it was easy enough to express $\displaystyle \mathbb{E}[X]$ in terms of the CDF and the result followed from there...
Ok.

But for the discrete case I'm a little stuck as to how to start...

$\displaystyle \mathbb{E}[X] = \sum_{i=0}^{{\color{red}\infty}}x_iP(X=x_i)$
$\displaystyle \sum_{n=1}^\infty\mathbb{P}(X\geq n) = \sum_{n=1}^\infty\sum_{i=n}^m\mathbb{P}(X=x_i)$

hmmmm....

ay help would be much appreciated
Hmm that's not really correct :s

It is not specified whether X have a continuous or a discrete distribution. And if it has integer values or not.

The problem with what you've done is that the $\displaystyle x_i$ have to be ordered, that is to say that they have to be in an increasing order.
Why ? Because you're considering $\displaystyle X\geq n$. This means that you consider all the $\displaystyle x_i$ that are $\displaystyle \geq n$.
By summing over the i, you don't include this important information.

You should do something like :
$\displaystyle \mathbb{P}(X\geq n)=\sum_{x \geq n} \mathbb{P}(X=x)$ (it's a summation over the x)

FYI, discrete distributions can also have a cdf.

I find your problem quite confusing... Is it really the way it has been posed ?

3. Yeah, it's posed exactly like that... I have reason to think the question setter may have only intended the continuous case to be considered, but I'd like to see how to show it for the discrete case too, if it's true.

I know discrete distributions can have a cdf, the problem is just how to express the expectation in terms of it if it's not specified that X can only take integer values.

I don't understand why you changed the summation to go to infinity for the expectation of a discrete distribution? I know there could be a (countably) infinite number of outcomes, but it could also be finite...

I see what you mean about ordering and summing over i though, thanks... But now the problem seems even more complicated! :S

4. Originally Posted by RanDom
Yeah, it's posed exactly like that... I have reason to think the question setter may have only intended the continuous case to be considered, but I'd like to see how to show it for the discrete case too, if it's true.

I know discrete distributions can have a cdf, the problem is just how to express the expectation in terms of it if it's not specified that X can only take integer values.

I don't understand why you changed the summation to go to infinity for the expectation of a discrete distribution? I know there could be a (countably) infinite number of outcomes, but it could also be finite...
But finite is "included" in infinite.

It it's finite, you can say that there exists N, such that for any n>N, $\displaystyle \mathbb{P}(X=n)=0$
And you would then have a finite summation.

While if you start with a finite summation, you don't take into account the infinite case !!!

I see what you mean about ordering and summing over i though, thanks... But now the problem seems even more complicated! :S
Not that much, if X has integer values (and not rational, or any other countable set)
I can try to find if the formula is correct for integer values, but it'd take some time, because I'm a little busy :s
Anyway, you can try it too. Think about reversing summation order.

5. But finite is "included" in infinite.

It it's finite, you can say that there exists N, such that for any n>N,
And you would then have a finite summation.

While if you start with a finite summation, you don't take into account the infinite case !!!
Ahhh, very good.... I meant for m to be allowed to be infinity but this is a better way to put it.

I don't know if X has integer-only values or not. I might try it as an exercise anyway though & ask the question setter about it when I ask him if he meant for X to be only continuous.

Thank you for your help

6. Yahoo ! I found it if $\displaystyle X\geq 0$, with integer values !

$\displaystyle \begin{gathered} \sum_{n=1}^\infty \mathbb{P}(X\geq n)\leq\mathbb{E}(X)\leq 1+\sum_{n=1}^\infty \mathbb{P}(X\geq n) \\ \Leftrightarrow \\ 0\leq \mathbb{E}(X)-\sum_{n=1}^\infty \mathbb{P}(X\geq n)\leq 1 \end{gathered}$

Now let's consider $\displaystyle p=\mathbb{E}(X)-\sum_{n=1}^\infty \mathbb{P}(X\geq n)$ (yes, calling it p was not RanDom ^^)

Note that $\displaystyle \mathbb{E}(X)=\sum_{n=0}^\infty n\mathbb{P}(X=n)=0 \cdot \mathbb{P}(X=0)+\sum_{n=1}^\infty n\mathbb{P}(X=n)=\sum_{n=1}^\infty n\mathbb{P}(X=n)$

Then... :
\displaystyle \begin{aligned} S &=\sum_{n=1}^\infty \mathbb{P}(X\geq n) \\ &=\sum_{n=1}^\infty \sum_{k=n}^\infty \mathbb{P}(X=k) \\ &=\sum_{k=1}^\infty \sum_{n=1}^k \mathbb{P}(X=k) \quad (\star) \\ &=\sum_{k=1}^\infty \mathbb{P}(X=k) \sum_{n=1}^k 1 \quad (\star\star) \end{aligned}

$\displaystyle (\star)$ :
Spoiler:
that's by reversing the summation order.
The range for k is 1 to infinity, since it can be equal to n, which can be equal to 1,2,...
The range for n will be {1,2,...,k}, because $\displaystyle k \geq n \Leftrightarrow n\geq k$ (yes, it sounds stupid to write this, but maybe it helps you visualize)

$\displaystyle (\star\star)$ :
Spoiler:
because $\displaystyle \mathbb{P}(X=k)$ doesn't depend on the indice of summation, n.

\displaystyle \begin{aligned} S &=\sum_{k=1}^\infty (k-1)\mathbb{P}(X=k) \\ &=\sum_{k=1}^\infty k\mathbb{P}(X=k)-\sum_{k=1}^\infty \mathbb{P}(X=k) \\ &=\mathbb{E}(X)-\sum_{k=1}^\infty \mathbb{P}(X=k) \end{aligned}

Therefore, $\displaystyle p=\mathbb{E}(X)-\mathbb{E}(X)+\sum_{k=1}^\infty \mathbb{P}(X=k)=\boxed{\sum_{k=1}^\infty \mathbb{P}(X=k)}$
But if you think more precisely about it, we then have :
$\displaystyle p=1-\mathbb{P}(X=0)$

Since $\displaystyle \mathbb{P}(X=0)$ is a probability, that is $\displaystyle 0\leq\mathbb{P}(X=0)\leq 1 \Longleftrightarrow 0\leq p\leq 1$
and you're done

7. Nice one! Well done