Stochastic_Jimmy
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- Jan 10, 2013
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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!
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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