PMF of Sum of Discrete Random Variables

[Stochastic_Jimmy]

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!

 

[DrPhil]

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..]

 

[Stochastic_Jimmy]

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.​

 

[DrPhil]

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

 

[Stochastic_Jimmy]

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!

 

[DrPhil]

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.

 

[Stochastic_Jimmy]

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.

Thanks DrPhil! Much appreciated.