Let $X_1$ and $X_2$ be two independent and identically distributed random variables having geometric distribution. The pmf is of the form $q^xp$, $x=0,1,2,...$ to infinity. I have to show $X_1=x_1|X_1+X_2=t$ has uniform distribution. I took $T=X_1+X_2$ and found the probability $P[T=t]$ which came out to be $(t+1)p^2q^t$. The required conditional distribution then had the pmf $\frac{1}{t+1}$. But if the distribution is uniform, the pmf should be a constant, am I right?
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No, that’s the conditional sign. I’m editing it to make it clearer, sorry. – AP _ Nov 29 '19 at 12:33
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In this context $\frac1{t+1}$ is a constant. – drhab Nov 29 '19 at 13:50
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The distribution is uniform on $\{0,1,2,...,t\}$.
Kavi Rama Murthy
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Can you explain further? I only know that the discrete uniform distribution has a pmf $\frac{1}{N}$ with x varying from 0 to N. – AP _ Nov 29 '19 at 12:36
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In this case $X_1$ cannot take values larger than $t$ because of the conditioning. It has uniform distribution with $N=t+1$. – Kavi Rama Murthy Nov 29 '19 at 12:53
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