# i.i.d. random vectors [duplicate]

If $$(X_1,Y_1), (X_2,Y_2)$$ are independent random vectors having the same joint distribution function $$F$$, then is it correct to say:

1. $$E(X_1)=E(X_2)$$ and $$E(Y_1)=E(Y_2)$$ (the same for variance);
2. Both, $$X_1$$ and $$Y_1$$, are independent of $$X_2$$ and $$Y_2$$ .

Please argue the answer.

This question arose when I read the equality in my book:

$$2Cov(X_1,Y_1)=E[(X_1-X_2)(Y_1-Y_2)]$$.

Thanks in advance.

## marked as duplicate by whuber♦ self-study StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 16 at 18:17

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• Part (1) appears to be trivial, because you ask whether a distribution function determines its moments: it does by the very definitions of distribution and moment. Therefore I have identified duplicates of your question (2). – whuber May 16 at 18:19