1
$\begingroup$

Conditions

1) Two random variables X, Y are independent and they follow binomial distribution with probability of success p.

2) X's mean is 2 and so does Y.

The question is to find the CDF of Z=min(X,Y)

I have no idea how to solve this question. I'm only familiar with continuous random variables not discrete ones. Any help would be appreciated.

  • 0
    Binomial with how many trials? – 2017-01-08
  • 0
    Note that $$F_Z(z)=P(Z\leqslant z)=1-P(Z>z)=1-P(X>z)P(Y>z)=1-(1-F_X(z))(1-F_Y(z))$$ hence $$F_Z(z)=F_X(z)+F_Y(z)-F_X(z)F_Y(z)$$ – 2017-01-08

1 Answers 1

1

$X \sim Bin(n,p)$, $Y \sim Bin(m, p)$

$np=2$, $mp=2$, hence $n=m=\frac2p$

if $z \in \mathbb{Z}$

\begin{align} P(Z \leq z) &= 1-P(Z > z)\\ &=1- P(\min(X,Y)>z) \\ &=1-P(X>z)P(Y>z), \text{ where I have used independence} \\ &= 1 - P(X>z)^2 \\ &= 1-\left(\sum_{i=z+1}^{\frac2p}\begin{pmatrix} \frac2p \\ i \end{pmatrix}p^i(1-p^{\frac2p-1})\right)^2 \end{align}