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Let X~Bernoulli$(\theta)$ and Y~Geometric$(\theta)$ where X and Y are independent. Let Z = X + Y. What is the probability function of Z?

My thoughts are:

Let Y be the number of failures until the first success.

$p_Z(0) = p_X(0)p_Y(0) = (1 - \theta)(1 - \theta)^0\theta = (1-\theta)\theta$ $p_Z(1) = p_X(0)p_Y(1) + p_X(1)p_Y(0) = (1 - \theta)(1 - \theta)^1\theta + \theta(1 - \theta)^0\theta= (1-\theta)^2\theta + \theta^2$

and for z >= 2,

$p_Z(z) = p_X(0)p_Y(z)+p_X(1)p_Y(z-1) = (1-\theta)(1-\theta)^z\theta + \theta(1-\theta)^{z-1}\theta$

Can someone tell me if I'm on the right track?

EggHead
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  • I find 2 definitions of Geometric distribution on http://en.wikipedia.org/wiki/Geometric_distribution. I think you are using the first one. Notice that in that case $p_{Y}\left(z\right)=\left(1-\theta\right)^{z-1}\theta$ and not $\left(1-\theta\right)^{z-1}$. Do not forget to look for $p_{Z}\left(0\right)$ apart. – drhab Oct 17 '13 at 22:29
  • Yes, there are actually 2 definitions. One refers to the number of tails (failures) before a head (success). The other refers to the number of tosses to get the first head (success). – EggHead Oct 18 '13 at 15:05
  • With wich one are you dealing here? – drhab Oct 18 '13 at 15:08
  • Hi drhab, Y is number of failures until first success. – EggHead Oct 18 '13 at 16:23
  • Then your present expressions for $p_{Z}\left(z\right)$ are formally correct. You can write it a bit shorter as. $\left(1-\theta\right)^{z+1}\theta+\left(1-\theta\right)^{z-1}\theta^{2}$. The case $z=1$ is not a special one. You get that result also if you substitute $z=1$ in the last equation meant for $z>1$ – drhab Oct 18 '13 at 18:50

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