Sufficient conditions for mutual independence

Take $n$ random variables $X_1,\ldots, X_n$ and suppose that $$X_i\perp X_1,\ldots, X_{i-1}, X_{i+1},\ldots, X_n \quad \forall i \in \{1,\ldots,n\} \tag{\star}$$ Does this imply that $X_1,\ldots, X_n$ are mutually independent?

I believe yes but I want to double check with you. Here's my proof where $f_{\cdot}$ denotes probability distribution or pdf.

• By $(\star)$ $X_1\perp (X_2,\ldots, X_n)$. Hence, $f_{X_1,\ldots, X_n} =f_{X_1}\times f_{X_2,\ldots, X_n}$

• By $(\star)$ $X_2\perp (X_1,X_3,\ldots, X_n)$ which implies $X_2\perp (X_3,\ldots, X_n)$ Hence, $f_{X_1,\ldots, X_n}=f_{X_1}\times f_{X_2}\times f_{X_3,\ldots, X_n}$

• By $(\star)$ $X_3\perp (X_1,X_2,X_4,\ldots, X_n)$ which implies $X_3\perp (X_4,\ldots, X_n)$ Hence, $f_{X_1,\ldots, X_n}=f_{X_1}\times f_{X_2}\times f_{X_3}\times f_{X_4,\ldots, X_n}$

• ... $f_{X_1,\ldots, X_n}=\prod_{i=1}^n f_{X_i}$ which means mutual independence.

• Note that $$\prod_{i=1}^n f_{X_i}$$ looks different from $$\Pi_{i=1}^n f_{X_i}$$ and $\prod_{i=1}^n f_{X_i}$ looks different from $\Pi_{i=1}^n f_{X_i}.$ I edited accordingly. – Michael Hardy Feb 7 '18 at 18:31
• What's the mathematical difference? – STF Feb 7 '18 at 18:39
• It's a difference in typesetting. Somewhat akin to correct spelling. – Michael Hardy Feb 7 '18 at 19:48