00:01
First of all, the region of the integration is this.
00:07
Sorry.
00:10
So this is the line y equal to x.
00:13
So it's given that 0 less than or equal to y less than x less than infinity.
00:24
So it will be this region.
00:30
So now, the joint mgf, fxy of t1 comma t2 is given by expectation of t part t1x plus t2y.
00:43
This is integration.
00:45
I'll choose a typical.
00:47
Y so the entry for x is equal to y and the exit for x is infinity so that means the coordinates of y is 0 to infinity but x will go from y to infinity e part t1x plus t to y into y per minus x d x d y so that means we can write this as you can write this integral as y 0 to infinity a part t to y into y into integration of x going from y to infinity e part t1 minus 1 of x d x d x d y so let's do the inner integration first so this inner integration will be e part t1 minus 1 into x divided by t1 minus 1 now substitute the upper and lower limits now here we need to rest it that t1 minus 1 is negative so once you substitute up unlimited is 0 minus 1 by t1 minus 1 basically it is 1 by 1 minus t 1 so that means the outer integral now becomes integration of y 0 to infinity y e power t 2 i into e power t 1 i'm sorry it's not 1 it is t 1 power t 1 minus 1 i i'm sorry it's not 1 it is t 1 2 1 minus 1 1 1 1 2 2 2 in the numerator it is minus of p power t 1 minus 1 into y so it will be e power t 1 minus 1 into y divided by 1 minus t 1 so divided by 1 minus t 1 d y so that means your mgf will be mgf will be mxy of t 1 t 1 t 2 is equal to 1 by 1 minus t 1 integration of 0 to infinity y e power t 1 plus t 2 minus 1 y d y and we have a condition that t 1 is less than 1 now using integer by parts you have y into integration of e part t1 t1 plus t2 minus 1 in the part t 1 plus t 2 minus 1 of y divided by t 1 2 2 minus 2 minus 1 minus difference of y is 1 again we need to integrate this it is e part t 1 plus 2 minus 1 of y divided by t 1 2 minus 2 minus 1 the whole square now substitute the limits 0 and infinity so when you substitute the limits 0 and infinity so 1 by one minus t one so upper limits will be zero because again we need to assume that even because t2 is less than one so upper says zero and the lower one will be this is zero but that will be minus one by t1 plus t2 minus one the whole square that's all so we got the mgf so mxy of t1 comma t2 is equal to one by one minus t1 into t1 plus t2 minus one the whole square where t1 is less than one and t1 is also less than one.
04:06
This is the mgf of the i mean the joint mgf of the two random variables x and y now to find the individual mgfs what we need to do is mx of t1 is obtained by substituting t2 is zero so substitute t2 is zero in the above it is one by one minus t1 whole cube and similarly to find the mgf of the random variable y put t1 is zero above could t1 is zero so it will become one by one minus t2 the whole square so these are the individual mgf if you multiply these two mgf are you getting this mgf not at all not at all x and y are not independent random variables because the product of this mgf and this mgf is not same as this mgf x and y are not independent so x and y are not independent now the next question that was asked is variance of x plus y okay so now we have now first what we do is let us find expectation of x square expectation of x expectation of y square expectation of y expectation of y expectation of y expectation of x y using the joint mgf to find expectation of x square and to find expectation of x we have to use the mgf after substituting t2 is zero that means this one this is the mg of the random variable x so mx of t1 if you differentiate with respect to t1 d by d t1 so what is the derivative of that it is minus 3 by 1 minus t 1 to the power of 4 into chain rule minus 1 so it's 3 by 1 minus t 1 to the power of 4 now substitute t 1 is 0 so you get expectation of x s one more time differentiate d square max of t1 divided by d t2 squared will be minus 12 by 1 minus t 1 to the power of 5 into chain rule that is negative 1.
06:14
Now again substitute t1 is 0 that will be 12 and that is expectation of x square.
06:20
So expectation of x square is 12.
06:22
Next let's go to the random variable y, the mjf of y, which is 1 by 1 minus t 2 whole square.
06:28
1 by 1 minus t 2 whole square is m y of t 2 is 1 by 1 minus t 2 whole square.
06:36
D m y of t 2 divided by d t 2 is equal to minus 2 by 1 minus t 2 the whole cube into negative 1 that is 2 by minus t 2 now substitute t is 0 expectation of y is 2 now one more time differentiate d square m y of t 2 divided by d t 2 square is equal to minus 6 by 1 minus t 2 whole power 4 into minus 1 and substitute t2 is 0 it will be 6.
07:08
So expectation of y square is equal to 6.
07:11
Now what is variance of x? variance of x is, variance of x is expectation of x square, that is 12 minus expectation of x whole square that is 9.
07:22
So it's 3.
07:23
Variance of x is 3...