00:03
We have been asked to check whether this function is a valid joint density function.
00:13
So in order to verify that all that we need to do, we need to do a double integration over the function with respect to both x and y.
00:22
And if the answer is equal to 1, then this function is a valid joint density function.
00:31
Now, so we're going to do our double integration over the function.
00:34
And i'm just going to treat this one first.
00:38
So with this one first, we are going to have integer and then know that because i'm dealing with respect to x, y is assumed a constant.
00:49
So we have 0 .1 and then we'll split to x would be 0 .5 and then e raise the power 0 .5, x minus 0 .2, y for x by 1 ,000, for x by is infinity to zero the y.
01:09
Now when you simplify the above, we are having 0 .1 minus 0 .5, and then when x is infinity, this whole function goes to 0.
01:24
And when x is 0, we have e minus 0 .2.
01:30
Y to y, now when we try to integrate over y then we are also going to get because this one is a constant we can just pull it out and then this is we'll go off so we are having infinity 0 um 0 .1 divided by 0 .5 and then e minus 0 .2 y do i and when you integrate with respect to y you should be able to get 0 .1 divided 0 .1 by 0 .2 so 0 .5 times 1 divided by minus 0 .2 e minus 0 .2 and then values of x is what infinity and then 0.
02:28
So when you simplify again you are going to get 0 .1 divided by minus this time that will give you 0 .1.
02:41
And then when y is infinity you are having zero and when y is zero the resulting answer is one so when we simplify this one then a thought is what to one so we have been able to verify that this function is indeed a joint density function now the next question said that we should be able to find the probability of y greater than one so probability of y greater or equal to one and then i always say that this one because it's a joint density function and they do have the corresponding um x this can also be written as y greater than one and then x greater or equal to zero so what i'm going to do we need to integrate over y values from zero to one and then x values from infinity to zero and then you have 0 .1 and then you have 0 .1 e minus 0 .5 x minus 0 .2 y the x the x the y so we're just going to treat first with the x values so we are having infinity 1 and then 0 .1 so when you integrate this one with respect to x we are having minus 0 .5 e minus 0 .5 x minus 0 .5 x minus 0 .5 x minus 0 .2 y and then for values of x being infinity to 0 and then y now when you try to simplify the whole of this you should be able to get infinity 1 and then 0 .1 minus divided by 0 .5 you have a negative there negative sign there already now when x is infinity this whole thing will be 0 and then when x is infinity this whole thing will be 0 and then when x is zero this whole thing we wrote e minus 0 .2y and i want you simplify this whole thing again we should be able to get 0 .1 divided by 0 .5 because this and this will cancel we have e minus 0 .2 y the y the y so when i try to integrate this one with respect to we should be able to get 0 .1 divided by 0 .5 as always there.
05:23
So times we have 1 divided by minus 0 .2 and then e minus 0 .2.
05:33
Y for x values infinity to 1.
05:37
And then this whole thing can also be simplified as 0 .1 divided by minus 0 .1.
05:45
Now when y is infinity we have zero but when y is one the resulting answer is what 0 .1 so 0 .8187.
06:01
So when it's simplified the whole of this you are getting minus 1 minus 0 minus 0 .817.
06:12
So the final answer is 0 .817.
06:22
The next question says that we can find the probability of what? x less than equal to 2 and then y less than equal to so just integrating over x and then because this is the highest value of y, so we have 40 and then zero and then the highest value of x is what two and you have zero and there are 0 .1 e minus 0 .5 x minus 0 .2 .2y the x the x the y.
07:04
So i'm just going to treat for respect to x as we usually do first.
07:10
So we should be able to get infinity 4 0 and then when you integrate this one with respect to x the y is treated as a constant so we are having 0 .1 minus 0 .5 e minus 0 .5 x minus 0 .2y minus 0 .2 y and then for x values from 2 to 0d1 now when we try to simplify this whole of this you have 4 here 0 .0 .1 minus 0 .5 now when x is 2 we have e minus 1 minus 0 .2 y and then when x is 0 we have e minus 0 .2y and then the y now i don't want to simplify this whole thing so this whole thing can be simplified as 4 0 .1 minus 0 .5 and then this whole thing thing i'm so e minus 1 because of this one can write e minus 0 .2y minus e minus 0 .2y and this whole again also be simplified as what four zero zero point one minus and then we can say e minus 1 minus 1 because i'm trying to factor this whole thing out so e minus 0 .2 y the y.
08:55
So i'm just trying to integrate this one with respect to why.
08:59
This is a constant.
09:00
This is also a constant.
09:01
So it is same as what.
09:04
0 .1 minus negative 0 .5 e minus 1 minus 1 and then when we integrate this one with respect to why we have 1 divided by minus 0.
09:19
0 .2 e minus 0 .2 y for x values 2 so is 4 and then 0 so when trying to simplify the whole of this we should be able to get 0 .1 divided by so i'm just multiplying this and this we are having 0 .1 and then we have a 0 .1 and then we have e minus 1 minus 1 and then when x so when y is 4 when y is 4 when y is 4 when y is 4 we are having e minus 0 0 and then when y is 0 we have 1 there now when you use a calculator to simplify the whole of this thing e sorry e raised of power minus one minus one times e raise of power 0 .8 minus 1 the answer is equals to what 0 .3481 so the probability for this whole thing is what 0 .384 so the probability for this whole thing is what 0 .384 and then the next one we have been asked to find the expected values of x and y so i always say that the expected value which can be written like this is the same as saying integrating the function over x and y and then multiplying because you are finding the expectation of x you just multiply by x and then you find the x the y so what are you are trying to say that this whole thing can be written as sort of infinity zero infinity zero 0 .1e minus 0 .5 x minus 0 .2 y sorry x the x the y now i'm just trying to integrate this whole thing with respect to x so this whole thing would respect to x but because you can see that this can be integrated by using integration by part so we have integration by parts here integration by parts okay yeah integration by part so we see if the function is what um if the function is what um um you dv in the same as what u v minus into bracket infinity sign u okay so from this function let our u equal to x so the u is the same as what the x now let our v equal to what the v equal to 0 .1 e minus 0 .5 x 0 .2y so if you integrate this one with respect to v rather v equal to 0 .1 minus 0 .5 e minus 0 .5 x minus 0 .2y so what do you do the whole of this one is what by multiplying this function by that because we have it here so you have what 0 .1 minus 0 .5 x e minus 0 .5 x.
13:41
E minus 0 .5 x minus 0 .2 y.
13:44
And then the values is what infinity to 0 minus the integration of what this is 1.
13:51
So you have infinity to 0 and then we have 0 .1 minus 0 .5 e 0 .5 x minus 0 .2y then x the y now when you evaluate this one this whole thing at infinity is equal to what 0 and then when it's zero when x is infinity is 0 when x is 0 is 0 is 0 is so we are still left with this one for us to integrate this one with respect to x so it should be the same as infinity 0 so we have a negative sign already here because of this one.
14:44
So 0 .1, 0 .5, and then e, 0 .5, x minus 0 .2 .2y, the x, the y.
14:58
So we are just trying to integrate this one, infinity, 0.
15:03
We are just trying to integrate this one with respect to x.
15:05
So this and this will multiply this and this will multiply and then it will be what 0 .1 0 .5 so when you integrate this one with respect to x we are having what times 1 divided by minus 0 .5 and then you have e 0 .5 x minus 0 .2 y for values infinity to 0 the y...