I have this variable $V = \min(X, Y)$, where $X$ and $Y$ are random variables geometrically distributed and independent. I want to compute the CDF. My attempt:
Let $k \in V(\Omega) = \mathbf{N}^*$. We have \begin{align*} \mathbb{P}(V \leq k) &= \mathbb{P}(\min(X,Y) \leq k) \\ &= 1 - \mathbb{P}(\min(X,Y) > k) \\ &= 1 - \mathbb{P}(X > k, Y > k) \\ &= 1 - \mathbb{P}(X > k) \times P(Y > k) \\ &= 1 - \mathbb{P}(X > k)^2 \\ &= 1 - (1 - p)^{2k} \end{align*} because $\mathbb{P}(X > k) = 1 - \mathbb{P}(X \leq k) = (1-p)^k$.
Is this correct? The solution given by the teacher was not this, and I feel this is right, as I want to get the PDF from the CDF as $$ \mathbb{P}(V = k) = \mathbb{P}(V \leq k) - \mathbb{P}(V \leq k - 1) = (1- p)^{2k - 2} - (1-p)^{2k} \,. $$ Please be as clear as possible.