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Suppose X and U are independent random variables with P(X = k) =$\frac 1 {N+1}$, k= 0, 1, 2, . . . ,N, and U having a uniform distribution on [0, 1]. Let Y = X + U.
a) For y ∈ R, find P(Y ≤ y).
b) Find the correlation coefficient between X and Y .

I got P(Y≤y)= $\frac y {N+1}$
So pmf of Y will be f(y)= $\frac 1 {N+1}$, y=1,2,...,N+1.
What should I take f(x,y) as?
Cov(X,Y) = E(XY)-E(X)E(Y).
E(XY) = $\sum$$\sum$$\frac {xy} {(N+1)^2}$ f(x,y), x=0,1,...,N;y=1,2,...,N+1

kris91
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  • Wrong title (Y is not discrete). – Did May 10 '13 at 09:56
  • y is discrete uniform. – kris91 May 10 '13 at 13:49
  • Nope. You might begin with this: what is really the distribution of Y? (By the way, how did you get the formula P(Y≤y)=y/(N+1) in your post?) – Did May 10 '13 at 14:06
  • http://math.stackexchange.com/questions/373656/cdf-of-sum-of-a-discrete-and-continuous-random-variable like this. – kris91 May 10 '13 at 14:22
  • And you still think that Y is discrete? Wow. – Did May 10 '13 at 14:28
  • So what is the distribution of Y? – kris91 May 10 '13 at 14:37
  • They give you that in the answer to the other question, no? And since you accepted the thing, presumably you understood it? Come! Think a little bit: X=0, 1, ..., or N, and 0≤U≤1 hence Y=X+U is such that... – Did May 10 '13 at 15:45
  • P(Y≤y)=P(X≤[y]-1)+ P(X=[y])*P(U≤y-[y]).From this I got P(Y≤y)=y/(N+1) – kris91 May 10 '13 at 16:07
  • And the distribution of Y is called... – Did May 11 '13 at 07:16
  • As far as know a discrete uniform distribution with pdf 1/(N+1) when the domain is 1,2,...,N+1 has this kind of cdf. – kris91 May 11 '13 at 09:31
  • OK, let us keep this simple: Y is NOT a discrete random variable, Y is a continuous random variable, Y takes values in the real interval [0,N+1] and the distribution of Y is uniform on the real interval [0,N+1]. Now I will post a solution with the fool's hope that you will read and ponder it and understand what is going on more than you did when accepting this other answer. – Did May 11 '13 at 09:42
  • And please change the title. – Did May 11 '13 at 09:51
  • and I did understand before accepting the other answer.I forgot about considering that Y could be continuous uniform. – kris91 May 11 '13 at 10:30

1 Answers1

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The distribution of $Y$ is uniform on the interval $[0,N+1]$. Thus, the density $f_Y$ of the distribution of $Y$ is defined by $f_Y(y)=\frac1{N+1}\mathbf 1_{0\leqslant y\leqslant N+1}$ for every real number $y$ and its CDF $F_Y$ is defined by $F_Y(y)=0$ for every real number $y\leqslant0$, $F_Y(y)=\frac1{N+1}y$ for every real number $0\leqslant y\leqslant N+1$ and $F_Y(y)=1$ for every real number $y\geqslant N+1$.

Now, $Y=X+U$ hence $\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X)+\mathrm{Cov}(X,U)$. The random variables $U$ and $X$ are independent hence $\mathrm{Cov}(X,U)=0$. The random variable $X$ is uniform on $\{0,1,\ldots,N\}$ hence $E[X]=\frac12N$ and $E[X^2]=\frac16N(2N+1)$.

Finally, $\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X)=E[X^2]-E[X]^2=\frac1{12}N(N+2)$.

Did
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