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Let $X_1,X_2$ be independent random variables.

Moreover $X_1,X_2$ are discrete uniform distributed({$1,...,N$})

We define:

$A:= X_1+X_2$

$B:= \min(X_1,X_2)$

Find joint-distribution of $A$ and $B$ and calculate covariance of $A$ and $B$.

I don't really have a clue how to find the joint-distribution of $A$ and $B$. So for this task I really need some help. Maybe you can give me a hint and I try to solve it then. I would edit this question with my attempt until I find the joint-distribution.

Edit: Let us start with the joint distribution. $P(A=a,B=b) =P(B=b|A=a)\cdot P(A=a)$.

In the answers below I saw that: $ P(A=a)=\frac{1}{N^2}\left\{ \begin{array}{lr}a-1& 2\leq a \leq N+1 \\ 2N+1-a & N+2\leq a \leq 2N \end{array} \right. $

I understand why this formula holds for $P(A=a)$ but only with the example. I don't know how we can show it. Moreover We have to find $P(B=b|A=a)$. I think a) is clear now. Will try to edit my attempt in a few days.

Edit for the second part: I will write $Cov$ for covariance. So we have to calculate $Cov(A,B)$.

We already know that: $Cov(A,B) = E(AB) - E(A)E(B) $ (expected value)

All we have to compute is the expected value for $AB,A,B$. Thanks to the user "mathemagical". I already know the value of $E(A), E(B)$. I even understand the rest of the answer except how we can calculate $E(Z)$.

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  • $\begingroup$ What does "Laplace-distributed ($\{1,2,\ldots N\}$)" mean? What's the significance of the $1,\ldots N$? The only Laplace distribution I know is the continuous distribution with PDF proportional to $e^{-|x|}.$ $\endgroup$ – spaceisdarkgreen Dec 17 '17 at 3:05
  • $\begingroup$ Sorry for the confusion. Laplace distribution is the another term for discrete uniform distribution. I will remark it. ! :) $\endgroup$ – Mugumble Dec 17 '17 at 3:10
  • $\begingroup$ en.wikipedia.org/wiki/Laplace_distribution The Laplace distribution describes a continuous random variable, and it is the difference between two independent exponential distributions. $\endgroup$ – Eric Fisher Dec 17 '17 at 3:20
  • $\begingroup$ Dear Eric Fisher, in my lecture my professor called the discrete uniform distribution the Laplace distribution. Don't ask me why. But I will edit it now, because I see that some people seem to be confused. Thank you for your reply :). $\endgroup$ – Mugumble Dec 17 '17 at 3:32
  • $\begingroup$ @Mugumble This is not standard (as far as I know) but from my cursory knowledge of history, Laplace seems as good a person as any other to name this after. $\endgroup$ – spaceisdarkgreen Dec 17 '17 at 3:41
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I am assuming that you mean the discrete uniform distribution that gives probability $\frac{1}{N}$ to each point. As the joint distribution is a bit gnarly, it seems easier to compute the covariance by computing the various pieces of $E(AB)-E(A)E(B)$.

Hints/Outline: (For easy typing, I have changed $X_1$ to $X$ and $X_2$ to $Y$).

For the covariance of $A$ and $B$ First find the distribution and expectation for A and B. Then do the same for $AB$.

From independence, the joint distribution of $(X,Y)$ is easy, with the probability associated with any $(x,y)$ pair given by $\frac{1}{N^2}$

Compute $E(A)$ and $E(B)$

Note that $A=X+Y$ takes values from 2 to 2N with probabilities as follows $$P(A=k)=\frac{1}{N^2}\left\{ \begin{array}{lr}k-1& 2\leq k \leq N+1 \\ 2N+1-k & N+2\leq k \leq 2N \end{array} \right.$$ (do you see why? The values taken by A for each X and Y when N=6 are shown in the table below, where X and Y are shown in the margins) $$\begin{array}{c|cccccc} 6&7&8&9&10&11&12\\ 5&6&7&8&9&10&11\\ 4& 5&6&7&8&9&10\\ 3& 4& 5&6&7&8&9\\ 2&3& 4& 5&6&7&8\\ 1&2&3&4& 5&6&7\\ \hline A&1&2&3&4& 5&6 \end{array}$$

This gives (derive ) $$E(A)=N+1$$ Note that $B=X$ when $X \le Y$ and $B=Y$ when $X\ge Y$ as in the example table below for $N=6$. $$\begin{array}{c|cccccc} 6&1&2&3&4&5&6\\ 5&1&2&3&4&5&5\\ 4&1&2&3&4&4&4\\ 3&1&2&3&3&3&3\\ 2&1&2&2&2&2&2\\\ 1&1&1&1&1&1&1\\ \hline B&1&2&3&4& 5&6 \end{array}$$ So (derive) $$P(B=k)=\frac{2N-2k+1}{N^2}$$ and $$E(B)=\frac{N(N+1)(2N+1)}{6N^2}$$

Compute $E(AB)$

Note that $$AB=\left\{ \begin{array}{lr} XY + Y^2& Y\le X\\XY+X^2&Y>X\end{array}\right.$$

$$AB=XY +\min(X^2,Y^2)=XY+Z$$

Now note that $E(XY)=E(X)E(Y)=\left(\frac{N+1}{2}\right)^2$ by independence. You can now finish the computation of $Cov(A,B)=E(XY)+E(Z)-E(A)E(B)$ by finding $E(Z)$ (you can recycle the distribution of $B$ for this: note that $P(Z=k^2)=P(B=k)$)

That then leads you to the covariance of A and B since you already have $E(A), E(B)$.

For the joint cumulative distribution of $A$ and $B$, use the geometry of the tables for $A$ and $B$ to note that there are 3 cases (and some meticulous computation for the probabilities in each of the 3 cases).

$$F(a,b)\equiv P(A\le a, B\le b)=\frac{1}{N^2} \left \{\begin{array}{lr} 2Nb-b^2& a\ge N+b\\ N^2-\frac{(2N-a)(2N-a+1)}{2}&a\le 2b\\ N^2-\frac{(2N-a)(2N-a+1)}{2} - \frac{(a-2b)(a-2b+1)}{2}& \mbox{otherwise} \end{array}\right.$$

If you want to find the probability mass function, then $$P(A=a,B=b)=F(a,b)-F(a-1,b)-F(a,b-1)+F(a,b)$$

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  • $\begingroup$ First of all: Thank you so much for your efforts! Your Assumption is right! So the discrete uniform distribution gives the probability $ \frac{1}{N} $. I see why $ P(A=k)$ is probability above ( with your example ) . To be honest I don't know how can I show this ( without the example ). Thanks to you I understand how to compute $E(A), E(B)$. Now I will read the rest of your answer and try to understand it! :) $\endgroup$ – Mugumble Dec 17 '17 at 13:54
  • $\begingroup$ @Mugumble Sure thing. I suggest that you derive the joint cumulative distribution function of A, B (you can figure out the geometries by starting hard at the table). Then the joint mass function is just a matter of differencing the cumulative. Simple. I am unable to draw figures because I'm using a phone (and won't have a computer for another month). But try drawing areas of what A < b, B< b and so on look like. Then try to see what's the area in which both are simultaneously true. You will get that 3 cases I mentioned. $\endgroup$ – Mathemagical Dec 17 '17 at 14:30
  • $\begingroup$ I understand your reasoning. Unfortunately I don't understand how we can compute $E(Z)$. I see that $P(Z=k^2) = P(B=k)$. So can you tell me if my following thought is correct: $ \sum_{i=1}^{N} i^2 \cdot P(Z=i^2) $ Now we can replace $P(Z=i^2) = P(B=i)$. Now we can replace $P(B=i)$ with $ \frac{2N-2i+1}{N^2}$. We have $ \sum_{i=1}^{N} i^2 \cdot \frac{2N-2i+1}{N^2}$, right? $\endgroup$ – Mugumble Dec 18 '17 at 20:12
  • $\begingroup$ @Mugumble That is exactly right. $\endgroup$ – Mathemagical Dec 18 '17 at 22:48
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Hint

I think the fastest progress will be to calculate the joint distribution directly (since you're asked for that anyway) and then use that to find the covariance via the standard formula.

You can get the joint using combinatorics. The simplest thing to directly compute is probably $$ P(B=b\mid A = a).$$ For instance conditional on $A=4,$ it is equally likely for $(X,Y)$ to be $(1,3)$ $(2,2)$ or $(3,1)$ so $B$ conditionally has a $1/3$ probability of being $2$ and a $2/3$ probability of being $1.$

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