Let U be uniformly distributed on the interval $(0, 2)$ and let V be an independent random variable which has a discrete uniform distribution on $\{0, 1, . . . , n\}$. i.e. $P\{V = i\} =\frac{1}{n+1}$ for $i = 0, 1, . . . , n.$ Find the cumulative distribution function of $X = U + V$ .
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Hint: In order to see what is going on, it is useful to give $n$ a concrete value, like $20$. Informally, we have a $21$-sided fair die with the numbers $0$ to $20$ written on its faces. We roll the die, and add to the result $V$ a number $U$, chosen from the interval $(0,2)$ with uniform distribution.
The result is our random variable $X$. We want to find the cumulative distribution function $F_X(x)$ of $X$. So we want to find, for any $x$, the probability that $X\le x$. Let us work with a particular value of $x$, like $x=8.74$.
Under what conditions is $X\le 8.74$? If $V$ takes on any of the values $0,1,2,3,4,5,6$, then for sure $X\le 8.74$. This is because with probability $1$, we have $U\le 2$.
The probability that $V\le 6$ is $\frac{7}{21}$.
But $X$ is also $\le 8.74$ if $V=7$, and $U\le 1.74$. The probability that this happens is $\frac{1}{21}\cdot\frac{1.74}{2}$.
Finally, $X$ is also $\le 8.74$ if $V=8$, and $U\le 0.74$. The probability that this happens is $\frac{1}{21}\cdot\frac{0.74}{2}$.
For the probability that $X\le 8.74$, add up. We get $$\Pr(X\le 8.74)=\frac{7}{21}+\frac{1}{21}\cdot\frac{1.74}{2}+\frac{1}{21}\cdot\frac{0.74}{2}.$$
The same reasoning will give you the cumulative distribution function $F_X(x)$ for any $x$ between $0$ and $n+2$. For completeness, one should add that $F_X(x)=0$ if $x\lt 0$, and $F_X(x)=1$ if $x\ge n+2$.
Let's start doing the general case. What did we do with $x=8.74$? First we took the integer part $\lfloor x\rfloor=\lfloor8.74\rfloor$. Then we backed up by $2$ to $\lfloor x\rfloor -2$. If $V\le \lfloor x\rfloor -2$, then for sure $X\le x$. But $X$ can also be $\le x$ in a couple of other ways: (i) if $V=\lfloor x\rfloor -1$ and $U$ is of suitable size or (ii) if $V=\lfloor x\rfloor$ and $U$ is of suitable size.
André Nicolas
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thank you. I had been trying to combine using sums of random variables but couldnt proceed because one is discrete and one continuous – kris91 Apr 27 '13 at 03:02
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You are welcome. It is more difficult, since one gets so little practice with these mixed distributions. – André Nicolas Apr 27 '13 at 03:12
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So basically the distribution of X is F(x)=(2x-1)/2(n+1) – kris91 May 05 '13 at 07:06
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The general expression involves the "floor" function. – André Nicolas May 05 '13 at 13:04
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they got cancelled.I will check once more. as in if V=[x]-1 then U≤x-[x]+1 – kris91 May 05 '13 at 13:38
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if I had to find correlation between X and V how do I do it?what do I take f(x,v) as?I can't just take it to be f(x)*g(v). – kris91 May 09 '13 at 05:58
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This really is a separate question that I couldn't deal with in a comment. Editing in comments is not easy, and the space is short. – André Nicolas May 09 '13 at 06:02
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ok.I will give the question.wait. – kris91 May 09 '13 at 06:03
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http://math.stackexchange.com/questions/386351/correlation-between-two-discrete-uniform-rvs – kris91 May 09 '13 at 06:22