00:01
Hello.
00:02
So in the following problem, we have been given that x and y are standard normal variables.
00:09
And we've been asked to consider these transformations that is u is x plus y and b is x minus y.
00:14
So and in the first part, we're supposed to find the function distribution, use the theory of distributions of functions to basically find the joint pdf of you and b.
00:27
And we're supposed to answer a couple questions in part a related.
00:31
To it such as what is a distribution of you, what is a distribution of we.
00:34
So the first thing i've noted is the pdf of x and y.
00:38
So since it's a normal function, this would be the pdf of x and this would be the pdf of y.
00:44
All right.
00:45
So let's go about working this question out.
00:48
We've been given something of the question related to a jacobian.
00:52
So let's find what that is.
00:55
Let's find what that is and work the problem out.
00:58
Another thing that the first part wants us to answer is, are u and v independent along with finding the distribution of u and v so given that we've been given u and v in terms of x and y but now let's try and find x and y in terms of u and v so x in terms of u and v and v so if you do that you get that x is actually u plus v over 2 and x is equal to u minus v over 2.
01:32
And therefore, if you do the jacobian, it's basically the partial of x with respect to you, partial of x with respect to v, partial of y with respect to you, and partial of y with respect to you, and partial of y with respect to v.
01:53
So this ends up being negative one half if you were to just calculate the derivatives of x with respect to you and you can do oh my bad i think i ended up writing this as x as well this should be y so we can find the partials from all these functions that are given and we get that it's negative half all right so in the next part what i'm going to do is i'm going to use the uh theory of distributions of functions so the what we can say is that the joint distribution of unv is the joint distribution of x and y times the absolute value of the jacobian.
02:44
We've been given that x and y are independent.
02:46
So it's safe to say that this is just the product of the individual pdfs of x and y, which we have established right at the start of the question.
02:56
So this ends up becoming 1 over 2 pi.
03:08
E to the negative half times x square plus y square times the absolute part of the jacobian which is one half so we basically have to find a joint distribution of u and v initially and what we have but what we have on the right hand side is terms and uh you know this this equation or this stuff on the right -hand side is in terms of x and y.
03:40
So how do we convert it to the terms of you and we? we've been given a relationship between x, x in terms of you and v and y in terms of you and we.
03:49
So if we plug those into x and y, you end up getting something on the lines of this.
03:57
So this would be 1 over 4 pi as the 2 and 2 multiplies into pi over here.
04:04
And this would essentially be e to the power negative one -fourth u -square plus v square.
04:12
I haven't elaborated all the steps in between, but basically if you just make the substitution, i'm just going to write here.
04:19
So take x equal to u plus v over 2 and y equal to u minus v over 2.
04:28
And if you make the substitution, this is what you'll get.
04:31
But the problem isn't complete here because now we need to establish, go on to find individual distributions of you and v.
04:41
So what i'm going to do is try to get this in similar terms of this.
04:50
So basically you want to show that this is somehow a product of the individual pdfs of you and v.
05:02
Right so the next steps would be trying to split this equation so what i'm going to say is this is one over two times square root of two square times square root of pi square right because four is just root two times root two times root two times root two and if you open those up it's basically root 2 times root 2 gives you 2 and then you have a root 2 square separately.
05:41
So you see i'm trying to get the root 2 pi in the denominator here in that form...