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Let $X$ and $Y$ have joint mass function

$f(j,k)=\frac {c(j+k)a^{j+k}}{j!k!}$, $j,k\geq 0$

where $a$ is a constant. Find $c$

This sum seems hard to to. How to complete this sum?

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We know that for joint density function, the integral over $X$ and $Y$ is equal to 1 –  Learner Mar 1 '13 at 4:41

2 Answers 2

up vote 1 down vote accepted

This is nothing but some calculation. Fix $k$ first,

When $k=0,\ \sum_{j=0}^{\infty} \frac{c(j+k)\cdot a^{j+k}}{j!\cdot k!}=\sum_{j=0}^{\infty} \frac{c\cdot j\cdot a^{j}}{j!}=ca\cdot e^a$

When $k\ge 1$, $$ \sum_{j=0}^{\infty} \frac{c(j+k)\cdot a^{j+k}}{j!\cdot k!}=\sum_{j=0}^{\infty} \frac{c\cdot j\cdot a^{j+k}}{j!\cdot k!}+\sum_{j=0}^{\infty} \frac{c\cdot k\cdot a^{j+k}}{j!\cdot k!}=\sum_{j=1}^{\infty} \frac{a^{j-1}}{(j-1)!}\cdot \frac{c\cdot a^{k+1}}{k!}+\sum_{j=0}^{\infty} \frac{a^{j}}{j!}\cdot \frac{c\cdot a^k}{(k-1)!}=ac\cdot e^a\cdot \frac{a^k}{k!}+ac\cdot e^a\cdot \frac{a^{k-1}}{(k-1)!} $$ $$ \sum_{k=1}^{\infty}(ac\cdot e^a\cdot \frac{a^k}{k!}+ac\cdot e^a\cdot \frac{a^{k-1}}{(k-1)!})=ac\cdot e^a \sum_{k=1}^{\infty} \big( \frac{a^k}{k!}+\frac{a^{k-1}}{(k-1)!}\big)=ac\cdot(2e^a-1) $$ Hence, $$\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\frac{c(j+k)\cdot a^{j+k}}{j!\cdot k!}=c\cdot2ae^{2a}=1\quad \Rightarrow\quad c=\big(2a\cdot e^{2a}\big)^{-1}$$ I hope the computation above is correct.

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Hint: It looks like $\frac c{j!k!} \frac d{da} a^{j+k+1}$ so binomial identities have a lot to say.

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