PMF of Sum of Discrete Random Variables

Stochastic_Jimmy

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I was hoping someone could tell me if I'm on the right track with this problem:

Let \(\displaystyle X \) and \(\displaystyle Y\) be independent random variables where \(\displaystyle X \sim Binomial(n,p) \) and \(\displaystyle Y\) is discrete uniform, equally likely to take on the values \(\displaystyle 0,1,2, \dots, n \). Find \(\displaystyle P(W = w) \), where \(\displaystyle W = X+Y \).

So we have \(\displaystyle P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \) for \(\displaystyle k \in [0,n] \) and \(\displaystyle P(Y = j) = \frac{1}{n+1} \) for \(\displaystyle j \in [0,n] \). Then:

\(\displaystyle P(W = w) = P(X+Y = w) = \sum_{w} P(X=k,Y=w-k) = \sum_{w} P(X=k)P(Y=w-k) \), by independence.

Since \(\displaystyle Y\) only takes on values between 0 and \(\displaystyle n\), we have for \(\displaystyle w-k \in[0,n] :\)

\(\displaystyle P(W=w) = \sum_{0 \leq w-k \leq n} \dfrac{ \binom{n}{k}p^k(1-p)^{n-k}}{n+1}. \)


Am I on the right track here? Thanks in advance for any comments!
 
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I was hoping someone could tell me if I'm on the right track with this problem:



So we have \(\displaystyle P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \) for \(\displaystyle k \in [0,n] \) and \(\displaystyle P(Y = j) = \frac{1}{n+1} \) for \(\displaystyle j \in [0,n] \). Then:

\(\displaystyle P(W = w) = P(X+Y = w) = \sum_{w} P(X=k,Y=w-k) = \sum_{w} P(X=k)P(Y=w-k) \), by independence.

Since \(\displaystyle Y\) only takes on values between 0 and \(\displaystyle n\), we have for \(\displaystyle w-k \in[0,n] :\)

\(\displaystyle P(W=w) = \sum_{0 \leq w-k \leq n} \dfrac{ \binom{n}{k}p^k(1-p)^{n-k}}{n+1}. \)


Am I on the right track here? Thanks in advance for any comments!
Sum over k, not w.
What are the limits of k for a given w?

[I have to go back and look again, since you edited while I was looking..]
 
Sum over k, not w.
What are the limits of k for a given w?

[I have to go back and look again, since you edited while I was looking..]

Thanks DrPhil. How's this:

Since \(\displaystyle Y\) only takes on values between 0 and \(\displaystyle n\), we need

\(\displaystyle 0 \leq w-k \leq n \implies w-n \leq k \leq w\)

So then: \(\displaystyle P(W=w) = \sum_{k=w-n}^w \dfrac{ \binom{n}{k}p^k(1-p)^{n-k}}{n+1}. \)

Thanks again for you help.​
 
Thanks DrPhil. How's this:

Since \(\displaystyle Y\) only takes on values between 0 and \(\displaystyle n\), we need

\(\displaystyle 0 \leq w-k \leq n \implies w-n \leq k \leq w\)

So then: \(\displaystyle P(W=w) = \sum_{k=w-n}^w \dfrac{ \binom{n}{k}p^k(1-p)^{n-k}}{n+1}. \)

Thanks again for you help.​
Good, but still needs more care with the summation limits.

\(\displaystyle k,j \in [0,n], \;\;\;\;\;\; w=k+j \in [0, 2n]\)

If w<n, the minimum value of k is 0 and the max is w,
...........and the corresponding range of j is w to 0

If w>n, the minimum value of k is w-n and the max is n,
...........and the corresponding range of j is n to w-n
 
Good, but still needs more care with the summation limits.

\(\displaystyle k,j \in [0,n], \;\;\;\;\;\; w=k+j \in [0, 2n]\)

If w<n, the minimum value of k is 0 and the max is w,
...........and the corresponding range of j is w to 0

If w>n, the minimum value of k is w-n and the max is n,
...........and the corresponding range of j is n to w-n

Would the PDF of \(\displaystyle W\) then be piecewise like this:

\(\displaystyle P(W = w) = \begin{cases}\sum_{k=0}^w \frac{ \binom{n}{k}p^k(1-p)^{n-k} }{n+1}, &\text{for }0 \leq w \leq n \\[10pt] \sum_{k=w-n}^w \frac{ \binom{n}{k}p^k(1-p)^{n-k} }{n+1}, &\text{for } n < w \leq 2n \\[10pt]
0, &\text{otherwise.}\end{cases} \)

Thanks again!
 
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Would the PDF of \(\displaystyle W\) then be piecewise like this:

\(\displaystyle P(W = w) = \begin{cases}\sum_{k=0}^w \frac{ \binom{n}{k}p^k(1-p)^{n-k} }{n+1}, &\text{for }0 \leq w \leq n \\[10pt] \sum_{k=w-n}^n \frac{ \binom{n}{k}p^k(1-p)^{n-k} }{n+1}, &\text{for } n < w \leq 2n \\[10pt]
0, &\text{otherwise.}\end{cases} \)

Thanks again!
Good. I changed the upper limit for the w>n case to be "n" instead of "w". The maximum number of terms in the sum is n+1.
 
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