sum of two discrete uniform random variables












0














Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.



My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$



I'm stuck here and don't know how to proceed. What other cases are needed to be considered?






probability







share|cite|improve this question






















  • Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
    – David G. Stork
    Nov 28 '18 at 5:08
















0














Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.



My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$



I'm stuck here and don't know how to proceed. What other cases are needed to be considered?






probability







share|cite|improve this question






















  • Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
    – David G. Stork
    Nov 28 '18 at 5:08














0












0








0







Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.



My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$



I'm stuck here and don't know how to proceed. What other cases are needed to be considered?






probability







share|cite|improve this question













Let $X$ be an integer chosen uniformly at random from the set ${1,2,...,n}$ and $Y$ be an independent integer chosen uniformly at random from the set ${1,2,...,m}$. Find the probability mass function of $X+Y$.



My attempt:
Without loss of generality suppose $n<m$. Possible values of $X+Y: 2,3,...,m+n.$ For $2leq kleq n$ using independence of $X$ and $Y$ we have
$P(X+Y=k)=sum_{i=2}^{n}P(X=k)P(Y=n-k)=frac{n-1}{mn}.$



I'm stuck here and don't know how to proceed. What other cases are needed to be considered?






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 28 '18 at 4:32









dxdydzdxdydz

1869




1869












  • Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
    – David G. Stork
    Nov 28 '18 at 5:08


















  • Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
    – David G. Stork
    Nov 28 '18 at 5:08
















Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08




Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, ldots, m+n$.
– David G. Stork
Nov 28 '18 at 5:08










1 Answer
1






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oldest

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0














You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$



Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$



To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$



So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).



In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.






share|cite|improve this answer























  • Is there an intuitive way to understand how we choose the boundaries for different cases?
    – dxdydz
    Nov 28 '18 at 5:53










  • I added that to the answer.
    – Alejandro Nasif Salum
    Nov 28 '18 at 6:01











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1 Answer
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1 Answer
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0














You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$



Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$



To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$



So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).



In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.






share|cite|improve this answer























  • Is there an intuitive way to understand how we choose the boundaries for different cases?
    – dxdydz
    Nov 28 '18 at 5:53










  • I added that to the answer.
    – Alejandro Nasif Salum
    Nov 28 '18 at 6:01
















0














You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$



Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$



To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$



So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).



In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.






share|cite|improve this answer























  • Is there an intuitive way to understand how we choose the boundaries for different cases?
    – dxdydz
    Nov 28 '18 at 5:53










  • I added that to the answer.
    – Alejandro Nasif Salum
    Nov 28 '18 at 6:01














0












0








0






You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$



Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$



To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$



So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).



In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.






share|cite|improve this answer














You have confused a little your variables and indexes. The sum for the case $2le kle n$ should be
$$sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$



Also consider the cases
$$n<kle m,quad 1le i le n$$
and
$$m<kle n+m,quad k-mle ile n.$$



To see this, think that $i$, the values that $X$ take, can be from $1$ to $n$, as long as the respective values for $Y$, that is $k-i$ are between $1$ and $m$. This implies:
$$1le k-i le m iff -m le i-k le -1 iff k-m le i le k-1.$$



So we have $ige 1$ and also $ige k-m$. This means that if $kle m$, $i$ can take the value $1$, but if $k>m$, then the minimum value for $i$ has to be $k-m$ (actually, this is also $1$ for $k=m+1$, but this does not contradicts what we said).



In the same way, $i$ has to be smaller or equal than $n$ and $k-1$, so if $kle n$, the maximum value for $i$ is $k-1$, but when $k>n$, the maximum value for $i$ is $n$. Combining these results, we get the three cases mentioned above.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 6:01

























answered Nov 28 '18 at 5:12









Alejandro Nasif SalumAlejandro Nasif Salum

4,409118




4,409118












  • Is there an intuitive way to understand how we choose the boundaries for different cases?
    – dxdydz
    Nov 28 '18 at 5:53










  • I added that to the answer.
    – Alejandro Nasif Salum
    Nov 28 '18 at 6:01


















  • Is there an intuitive way to understand how we choose the boundaries for different cases?
    – dxdydz
    Nov 28 '18 at 5:53










  • I added that to the answer.
    – Alejandro Nasif Salum
    Nov 28 '18 at 6:01
















Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53




Is there an intuitive way to understand how we choose the boundaries for different cases?
– dxdydz
Nov 28 '18 at 5:53












I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01




I added that to the answer.
– Alejandro Nasif Salum
Nov 28 '18 at 6:01


















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