We have discussed the distributions of random
variables. So, when we have a sample space and we are interested in certain characteristics
arising out of that experiment, such as, we are recording the heights of the individuals,
life of equipment, time taken by a sprinter to complete a 100 meter race, etcetera. So,
these are the examples of random variables. However, many times we may not be interested
directly in the same characteristic, but a function of that characteristic. Consider
measurements, where we are recording the errors in the actual measurement. So, the errors
may be negative or positive. However, it may turn out that, our losses are dependent upon
the absolute value of the error in the recording. That means, in place of the random variable
X, we may be interested in modulus of X. Suppose X denotes certain astronomical distances;
now, if the distances or the numbers are too large, we may be interested in a function
of X say log of X to make them scale to our level. In a similar way, sometimes we may
be interested in X square, we may be interested in say aX plus b or sine of X, etcetera. Now,
it is one thing to consider the characteristics of a function. So, in general, suppose I say
g X. We may be interested to look at expectation of g X, expectation of g square X, variance
of g X, etcetera. However, we may be interested in the actual distribution or the full probability
distribution of g X also. Now, in order to study the distribution of
g X, it is important to firstly ensure that g X is also a random variable. Recall that
the definition of a variable says that random variable is a real valued measurable function
defined on the sample space. Therefore, if g is also a measurable function on the real
line, then Y is equal to g X will also be a random variable. So, we start from this
result. Let X be a random variable defined on say omega, B, P. Let g be a measurable
function. So, g is a function from R to R. Then, g X is also a random variable. So, to
look at the proof of this, we must prove that the set of all those points, such that g X
omega is less than or equal to say y is a measurable set for every real y. Now, this
is the set X omega belongs to g inverse of minus
infinity to y. Now, if g is a measurable function, then g inverse minus infinity to y is a measurable
set. And therefore, X omega belonging to that – this is also a measurable set since X
is a random variable. So, since g is a measurable function, the set g inverse of minus infinity
to y is measurable set. And, since X is a random variable, the set
omega such that X omega belongs to g inverse minus infinity to y, is also measurable. So,
in general, we will be considering measurable functions of random
variable, so that to ensure that they are also random variable. And then, we can study
the probability distributions. Now, we have to ensure that a probability distribution
can be found. So, given a random variable X with cumulative distribution function, say
capital F X, the distribution of random variable Y is equal to g X, where g is measurable is
determined. So, now onwards, whenever we are considering the function g X, then g has to
be a measurable function. Let us consider the c d f of y. So, let me
use the notation say G y, that is, probability of Y less than or equal to y. Now, this is
equivalent to probability of g X less than or equal to y, which is equivalent to probability
of X belonging to the set g inverse minus infinity to y. Now, g inverse minus infinity
to y is a measurable set and the probability distribution of X is well-defined. Therefore,
this probability can be determined. Since g is measurable, g inverse minus infinity
to y is a measurable set. And, since the distribution of X is well-defined, G y is also well-defined.
So, the basic existence of the probability distribution of a function of random variable
is established. Now, let us look at the practical aspect of it; that means, how do we determine
the distribution of a function of random variable. This is a general approach. So, if we use
this c d f approach, that means we express the c d f of the function in terms of c d
f of X here. So, let us look at this. Let us consider a function say Y 1 is equal
to say aX plus b. So, if we are looking at the c d f of y 1, here a is a nonzero constant
and b is any real. So, this is probability of Y 1 less than or equal to say small of
y 1. Now, this is aX plus b less than or equal to y 1. Now, this can be expressed as probability
of X less than or equal to y 1 minus b by a if a is positive. And, it will become probability
of X greater than or equal to y 1 minus b by a if a is negative. So, notice here that
both of these are certain probabilities related to the random variable X. If the c d f of
X is well-defined, that means, the capital F X is known, then both of these probabilities
are also known; that means, this is equal to for example, F of X at y 1 minus b by a.
And, the second term we can write as 1 minus probability of X less than y 1 minus b by
a, which we can write as X less than or equal to this plus probability of X is equal to
y 1 minus b by a. So, this is equal to F X at y 1 minus b by a if a is positive; and,
1 minus F of X y 1 minus b by a plus probability of X is equal to y 1 minus b by a if a is
negative. Therefore, you can see that the c d f of y 1 is well-determined. Another thing
that we can note here is that if X is a continuous random variable, then this probability will
be 0. So, we will have only these terms here. Let us take another example say Y 2 is equal
to modulus of X. So, if we look at the c d f of Y 2 – now, notice here that if y 2
is a negative number, then this probability is going to be 0, because modulus of a random
variable is always a non-negative quantity. So, this is 0 if y 2 is less than 0. Now,
if y 2 is greater than or equal to 0, then we can express this as probability of minus
y 2 less than or equal to X less than or equal to y 2. So, this is equal to probability of
X less than or equal to y 2 minus probability of X less than minus y 2, which is nothing
but the c d f of X at the point y 2 minus the left-hand limit of the c d f at minus
y 2; or, we can write it as F X of y 2 minus F X at minus y 2 plus probability of X is
equal to minus y 2. Therefore, the c d f of y 2 is expressed as 0 if y 2 is less than
0 and it is F X of y 2 minus F X of minus y 2 plus probability of X equal to minus y
2 if y 2 is greater than or equal to 0. Here you can note that if y 2 is equal to 0, then
this term will vanish and this will reduce to probability of X is equal to 0. Also, if
X is a continuous random variable, then this term will vanish. Let us take say Y 3 is equal to X square.
Once again if I consider F of Y 3, then this is 0 if y 3 is less than 0. And, it is equal
to probability of Y 3 less than or equal to y 3 if y 3 is greater than or equal to 0.
Now, this quantity becomes probability of minus square root y 3 less than or equal to
X less than or equal to root of y 3. So, that is equal to F X of root y 3 minus F X of minus
root y 3 minus, that is, the left-hand limit at this point, which we can express as this
plus probability of X is equal to minus root y 3.
Let us consider say function Y 4 is maximum of X and 0. So, here F of Y 4 – once again
you can notice here that this random variable is also no negative; so, this will be 0 if
y 4 is negative. If y 4 is equal to 0, then this is simply probability of X less than
or equal to 0. And, if I consider y 4 positive, then it is the probability of X less than
or equal to y 4. So, that will be X less than 0 plus probability of 0 less than or equal
to X less than or equal to y 4. So, if I combine these terms, then this is equal to 0, if y
4 is less than 0; and, it is F X of y 4 if y 4 is greater than or equal to 0. You can
consider it as truncation of X at 0. So, this approach, where we find out the inverse in
each of the set g X less than or equal to c, that is, g inverse of minus infinity to
c and then expressing the c d f of the function g X in terms of the c d f of x. However, many
times the function may be quite complicated and we may not be able to express the inverse
image set in a proper way. So, we look at the particular approaches, when the random
variables are discrete or continuous. And, we may consider the special methods, for example,
if I have a one-to-one function or if I have two-to-one function, and if the random variable
is discrete, then we may express the probability of a point in terms of the inverse images
taken by the random variable. If we have a continuous random variable, then in place
of probability mass function, I will have a probability density function; then, there
is some method. So, let us develop these methods. In case X is a discrete random variable say
the random variable is taking values, say x i is belonging to some set script X; Y is
equal to g X, is a function. So, we consider that for y is equal to y j, g inverse of y
j could be x 1, x 2, x r say; that means, in general, suppose r inverse images are there,
then probability of Y is equal to y j – that will be equal to probability of X is equal
to x 1 plus probability X equal to x 2 plus probability X equal to x r, because what we
are doing is, we are writing it as probability of g X is equal to y j; that is probability
of X is equal to g inverse of y j. Now, what are the values of x i, which lead to g X is
equal to y j. So, we look at the set of inverse images and add all the probabilities.
As an example, consider probability X is equal to say minus 2, is equal to 1 by 5; probability
X is equal to minus 1, is equal to say 1 by 6; probability X equal to 0 is 1 by 5; probability
X equal to 1 is say 1 by 15; and, probability X is equal to 2 is say 11 by 30. Let us consider
say Y is equal to X square. Now, here X takes values minus 2, minus 1, 0, 1 and 2. So, this
Y will take values 0, 1 and 4. Now, probability of Y is equal to 0. Now, this has only 1 inverse
image, that is, probability X equal to 0. That will be 1 by 5. If we are looking at
probability Y is equal to 1, then there are two inverse images: X equal to minus 1 and
X equal to 1. So, the probability is 1 by 6 and 1 by 15, will be added up; we get 7
by 30. In a similar way, probability Y is equal to 4. Now, this will be probability
of X equal to minus 2 and probability X equal to 2. So, we will add 1 by 5 and 11 by 30,
which is leading to 17 by 30. So, the probability distribution of Y you can see here, is described
by probability Y is equal to 0, probability Y is equal to 1 and probability Y is equal
to 4. Now, this approach is not directly applicable when we have say mixture random variable or
if the function g X is such that the X may discrete, but g X may not be discrete; or,
X could be continuous, but g X may not be continuous. Let us take this case. Suppose X is a uniform
random variable on the interval minus 1 to 1. And, we consider the function say Y is
equal to maximum of X and 0. Now, you can see here that this distribution, this random
variable is not completely discrete nor continuous, because probability of Y is equal to 0 is
half. This is equivalent to probability X less than or equal to 0; that is equal to
half. But, if I look at probability Y is equal to half, etcetera, then that is 0, because
there after, the random variable is continuous. If X is non-negative, then Y becomes continuous
random variable. So, in this case, we can use the formula that we developed for the
c d f of Y 4, that is, maximum of X, 0. So, for y 4 less than 0, it is 0. And, thereafter,
it is F X of y 4. So, if we use this, we get probability of Y less than or equal to y;
it is equal to half for y is equal to 0 and it is equal to half plus y by 2 for 0 less
than y less than or equal to 1. And, of course, it is 1 for y greater than 1.
We have then density function here, that is, f Y y is equal to half for 0 less than y less
than 1. We have the weight half attached to probability Y is equal to 0; and, in the interval
0 to 1, we have a density function. So, it is an example of mixed random variable. Although
the random variable X is a continuous random variable, but the function of that is a mixture
random variable. We have certain conditions under which from a continuous random variable,
the function is also continuous random variable and the probability density function of the
given function Y is equal to g X can be determined using a formula. So, we state it in the following
theorem. Let X be a continuous random variable with
probability density function say f X X. Let y is equal to g x be a differentiable function
for all x and either g prime x is positive for all x or g prime x is negative for all
x. Then, Y is equal to g X, is a continuous random variable with
p d F given by f Y y is equal to f X g inverse y multiplied by the absolute value of d by
d y g inverse y over the range alpha to beta and 0 elsewhere, where alpha is the minimum
of the g minus infinity, g of plus infinity. And, beta is the maximum of g minis infinity,
g of plus infinity. Notice here the conditions. We are taking the function, such that the
function is differentiable everywhere and either the derivative is strictly positive
throughout the range or strictly negative. This ensures that the function is strictly
increasing or strictly decreasing. Also, it ensures that the function will be a one-to-one
function. In that case, there is a direct formula for the determination of the probability
density function of Y is equal to g X. It is described in terms of the density function
of X itself; that is, in place of X, we substitute g inverse y, which is uniquely determined
under the conditions given here. And, we multiply by absolute value of d x by d y, because if
y is equal to g x, then x is equal to g inverse y. And, this term is nothing but dx by dy.
Let us consider the proof of this. Let g prime x to be strictly positive for all x. Then,
g is strictly increasing; and so, it will be a one-to-one function. And, g inverse will
also be strictly increasing, that is, d by d y of g inverse y will be positive. And,
this will be differentiable also. So, if we consider the c d f of Y; and, another thing
is that the range we have to consider. If x is over certain range; suppose from minus
infinity to infinity in general; then, if g is an increasing function, then the minimum
value will be g of minus infinity and the maximum value will be g of plus infinity.
If g is a strictly decreasing function, then it will be reverse. So, here the range of
y will be from alpha to beta, where alpha and beta are defined like this. So, this is
equal to probability of g X less than or equal to y; that is, probability of X less than
or equal to g inverse y. This is ensured, because g is one-to-one function. So, g inverse
is a one-to-one function. Therefore, the regions g X less than or equal to y and X less than
or equal to g inverse y are equivalent. That means, whenever g X less than or equal to
y is satisfied, X less than or equal to g inverse y is also satisfied. So, this is nothing
but the c d f of X at g inverse y. Therefore, the density function of y will be determined
by differentiation of capital F Y. So, we get the probability density function
of Y is f Y is equal to f X g inverse y multiplied by d by d y of g inverse y. Since g inverse
y was increasing function, the derivative was positive. Therefore, this is equivalent
to absolute value of this. Now, in case g prime x is strictly negative, then g is strictly
decreasing and g inverse y will also be a strictly decreasing
function; that is, d by d y of g inverse y will be less than 0. So, when we consider
probability of Y less than or equal to y, this will be equivalent to X greater than
or equal to g inverse y. Since g is a decreasing function, here the event g X less than or
equal to y will be equivalent to X greater than or equal to g inverse y. So, which we
can write as 1 minus F X of g inverse y plus probability of X is equal to g inverse y.
Now, this term will be 0, because X is continuous. So, the c d f of y is determined in terms
of the c d f of X. So, if you differentiate, we get the p d f of y as minus f X g inverse
y into d by d y of g inverse y. Since d by d y of g inverse y is negative, minus of this
is absolute value. So, this is f X of g inverse y multiplied by d by d y of g inverse y. So,
you can see that in both the cases, the density function of y is determined as the density
function of X at the point X is equal to g inverse y multiplied by the absolute value
of d x by d y. So, this theorem is useful when the function g X is a strictly increasing
or strictly decreasing function. Let us take an example say the random variable
x is having a density 0 for x less than or equal to 0. It is equal to half for 0 less
than x less than or equal to 1. And, it is equal to 1 by 2 x square for 1 less than x
less than infinity. And, consider the function say Y is equal to 1 by X. So, here g x function
is 1 by x. So, g inverse function is also same. If I write this as y, y is equal to
1 by x. So, x will be equal to… So, if I look at this, this is a strictly decreasing
function; g is strictly decreasing and g inverse is also strictly decreasing. So, d by d y
of g inverse y is equal to minus 1 by y square. The density function of y then is determined
by the density function of x at g inverse y multiplied by the absolute value of the
d x by d y term. Now, here it is 0. So, it will remain 0. Whenever x is less than or
equal to 0, 1 by x is also less than or equal or 0. When it is half, this remains half multiplied
by 1 by y square. Now, the range 0 less than x less than 1 is translated to y is greater
than or equal to 1. And, in the third region, it is y square by 2 multiplied by 1 by y square.
When x is greater than 1, it will reduce to 0 less than y less than 1.
After simplification, this is equal to 0 for y less than or equal to 0; half for 0 less
than y less than 1; 1 by 2 y square for y greater than or equal to 1. Notice here that
these f X x and f Y y; they resemble. So, for x less than or equal to 0, it is 0; here
it is 0. Here it is 0 less than x less than or equal to 1; then it is half. And, when
x is greater than 1, it is 1 by 2 x square. So, that is also satisfied here, except that
equality at the end point. But, that hardly matters, because if I put y is equal to 1
here, the value is half and here the value at y is equal to 1 is half and here also;
actually at the end points, because it is a continuous random variable; the values will
be immaterial. So, basically, X and Y… X and 1 by X have the same distribution here. Let us look at another example. Say X follows
uniform 0, 1 and we define random variable U is equal to X divided by 1 plus X. Now again,
you can see here that this is a one-to-one function. In fact, if u is equal to x by 1
plus x, then the inverse function x is equal to u by 1 minus u. Let us look at the derivative
dx by du; that is equal to 1 by 1 minus u whole square. So, here the density function
of x is 1 for 0 less than u less than 1. So, the density function of u is obtained – 1
into 1 minus square. What will be the range? When this is 0 less than x less than 1…
So, when x is 0, this is 0; when x is 1, then this is half. It is a strictly increasing
function. So, 0 less than u less than half; and, it is 0 elsewhere.
Many times, the function g X may not be a one-to-one function; it may be a many one
function, for example, Y is equal to X square, Y is equal to modulus X. So, in that case,
we see that the region, that is, from r to r, the functions domain and range. So, what
we do, we look at the inverse image for a given y. And, if there are two inverse images,
then we split the region, that is, the domain of X into 2 disjoint regions, such that both
of them map g from each part of the domain to the full range. For example, you consider
Y is equal to X square. Now, from minus infinity to infinity, this maps to 0 to infinity. Now,
for a given Y, which is positive, I have two inverse images: minus square root y and plus
square root y. So, if I consider two portions of the domain, that is, minus infinity to
0 and 0 to infinity, both are mapped by this mapping to 0 to infinity. So, the idea here
is that in each part of the domain, the function will be one-to-one. That is, if you are considering
only one inverse image say root y or minus of root y, then the function is either increasing
or decreasing. We calculate the density in each region separately and add. This gives
the density function of the continuous random variable in case the function Y is equal to
g X is a many-valued function. So, let us look at the result here. Let X
be a continuous random variable with probability density function say f X x. Let y is equal
to g x be a differentiable function and assume that g prime x is continuous
and nonzero at all, but a finite number of values of x. Then, for every real number y, there exists
a positive integer n is equal to say n of y and real inverses say x 1 of y, x 2 of y,
x n of y such that g of x k of y is equal to y and g prime of x k of y is not 0 for
k equal to 1, 2 up to n of y. Or, there does not exist any X such that g x is equal to
y, g prime x is not 0 in which case we write n y is equal to 0. Then, Y is a continuous
random variable with p d f given by sigma f of x k y g prime x k y inverse; k
equal to 1 to n. This is if n is greater than 0; n means n y; and, it is equal to 0 if n
is 0. So, the idea is that we consider n separate regions of the domain such that each region
maps to the range of g. Calculate the density in each area and sum over all such areas.
That gives the density function of Y. So, let us look at application of this. I
am skipping the proof of this theorem. Let X follow say uniform distribution on minus
1 to 1. So, the density function is half for minus 1 less than or equal to x less than
or equal to 1. It is 0. Consider the function say Y is equal to modulus X. So, here for
a given y, we have g 1 of y is equal to minus y and g 2 y is equal to plus y. So, two inverse
images for a given y is there for y positive. If y is negative, n is equal to zero; that
means, there is no inverse image. And, the derivatives here you can see d by d y of this
is equal to minus 1 and d by d y of this is plus 1. So, if you take absolute values, then
both are 1. So, the density function of Y is the density function of X at minus y into
1 plus the density function at plus y into 1 for y greater than 0. If you want, you can
include equal to 0 also; that does not make any difference. And, it is equal to 0 for
y less than 0. So, this is half plus half; that is equal to 1.
Now, when we say y is greater than or equal to 0, here it will reduce to the region half
plus half for minus 1 to 1. So, it is becoming modulus of x between 0 to 1. And, it is equal
to 0 for y outside 0 to 1. So, you can see that Y follows uniform 0, 1. If X is minus
1 to 1 uniform distribution, then modulus of X is uniform distribution on the interval
0 to 1. Let us consider say X follows normal (0, 1)
distribution. So, the density function is 1 by root 2 pi e to the power minus x square
by 2. Consider the function say Y is equal to X square. So, now, for a given y, which
is non-negative, we have two inverse images; that is, g 1 of y is equal to minus root y
and g 2 of y is equal to plus root y. So, if I look at the derivatives, that will be
minus 1 by 2 root y or plus 1 by 2 root y. So, if I take absolute value, it is reducing
to 1 by 2 root y. For y less than 0, there is no inverse image. So, the density function
of y is the density at x equal to minus root y multiplied by 1 by 2 root y plus the density
function at plus root y multiplied by 1 by 2 root y. So, now, you see, the value will
be 1 by root 2 pi e to the power minus y by 2. And, in the second term also, same terms
will be coming, because minus root y and plus root y – both will give x square is equal
to y. So, 1 by 2 root y, 1 by 2 root y term is coming. So, it will become 2 times 1 by
2 root y. That is equal to 1 by 2 to the power half gamma half e to the power minus y by
2 into y to the power 1 by 2 minus 1 for y greater than 0. This is nothing but a gamma
distribution with r is equal to half and lambda is equal to half. So, you can see here the
square of a standard normal random variable is a gamma random variable.
Now, sometimes, it may happen that in place of finite number of inverse images, we have
infinite number of inverse images. That may happen in cases such as periodic functions,
which are trigonometric functions such as sine function or cos function, etcetera. So,
the theorem can be extended here. In place of a finite number of inverses, we have infinite
number of inverses. So, we again split the domain into infinite number of distinct regions
such that each of them is mapping to the full range of g. Calculate the density in each
region by the same formula, that is, F X of g inverse y multiplied by the absolute value
of d by d y of g inverse y and add in all the regions. Let us take one example here say f X x is
theta e to the power minus theta x for x greater than 0; that means, it is an exponential distribution
with parameter theta. Consider Y is equal to say sine of X. So, we consider sine inverse
y to be principal value. So, we consider two cases, because sine lies between minus 1 to
1. So, we consider case 0 to 1 and minus 1 to 0. So, if we take 0 to 1, then probability
that sine X is less than or equal to y can be expressed as probability of 0 less than
X less than or equal to sine inverse of y, where sine inverse is the principal value
plus probability of 2n minus 1 pi minus sine inverse y less than or equal to X less than
or equal to 2n pi plus sine inverse y; n is equal to 1 to infinity. So, this takes care
of all the infinite number of distinct regions, each of which map to the region sine X less
than or equal to y. Now, this probability using the exponential
density function, probability of X lying between 0 to sine inverse y is 1 minus e to the power
minus theta sine inverse y plus summation n is equal to 1 to infinity e to the power
minus theta 2n minus 1 pi minus sine inverse y minus e to the power minus theta 2n pi plus
sine inverse y. If we look at this series here, e to the power sine inverse y terms
can be separated out and the remaining terms become geometric sums and infinite geometric
series can be added. So, this is simplified to 1 plus e to the power minus theta pi plus
theta sine inverse y minus e to the power minus theta sine inverse y divided by 1 minus
e to the power minus 2 pi theta. In a similar way, if y between minus 1 to
0, we can split the regions and evaluate. So, after carrying out the calculations, the
density of Y can be obtained after differentiation as theta e to the power minus theta pi 1 minus
e to the power minus 2 theta pi inverse 1 minus y square to the power minus 1 by 2;
e to the power theta sine inverse y plus e to the power minus theta pi minus theta sine
inverse y. This is for minus 1 less than y less than 0. And, it is equal to theta 1 minus
e to the power minus 2 theta pi inverse 1 minus y square to the power minus 1 by 2 e
to the power minus theta sine inverse y plus e to the power minus theta pi plus theta sine
inverse y. This is for y between 0 and 1. And of course, it is 0 for all other values
of y, because y will lie between minus 1 to 1. So, if the transformation Y is equal to
g X is such that the random variable y is equal to g x is also continuous. Then, the
probability density function of Y can be determined in terms of density function of X. If the
function is one-to-one, there is a direct formula. If there is a many-one function,
then we have to split the domain into disjoint sets such that each part of the domain maps
to the full range. Calculate the inverse image in each of them and utilize that to find out
the density function of Y in each part separately and then sum.
There is one important result here, which connects all the continuous distributions,
which is known as the probability integral transform. This basically says that if X is
a continuous random variable with c d f, capital F; if we define say U is equal to F X of X,
then U is distributed on the uniform interval 0 to 1. So, this is known as probability integral
transform. The converse of this result is also true;
that is, if U is uniform 0, 1 and F is an absolutely continuous c d f, then X is equal
to F inverse of U, has c d f F X; basically, F here. Now, this is a very crucial result.
First thing is that it connects all the continuous random variables. So, if Y 1 is a continuous
random variable with some c d f F 1, Y 2 is a continuous random variable with some c d
f, then there exists a function g such that Y 2 is equal to say g of Y 1 and the distribution
of Y 2 will be given by this. So, basically, what we can do is that we can consider F 1
of Y 1, that is, say U. And then, we consider F 2 inverse of U; then, that will be this.
So, basically, F 2 inverse of F 1 Y 1 is equal to Y 2.
In the modern age of simulations, this result is quite useful. So, in some practical problem,
we may be interested to generate the values of a random variable, which is said having
say exponential distribution. So, we will use a pseudo-random number generator to generate
numbers uniformly between 1 to say some n. And then, we can consider division by n to
make it a uniform random variable between the values of the uniform random variable
on the interval 0 to 1. Now, if we are having the exponential distribution, the c d f of
that is known, that is, capital F is known, we take F inverse of that. So, suppose I consider
F x is equal 1 minus e to the power lambda x. So, if y is equal to 1 minus e to the power
minus lambda x, the inverse function can be obtained. So, minus lambda x is equal to log
of 1 minus y. So, x is equal to minus 1 by lambda log of 1 minus y. So, if y are the
uniform random variables on 0 to 1, then if we consider log of 1 minus y and minus 1 by
lambda, then x i’s will be exponential distributed random variables with parameter lambda.
These transformations play extremely important role in the simulation of random variables,
because we can use some pseudo-random number generator to generate uniformly distributed
random values. And now, for any other distributions, we make use of the transformations. So, especially
the probability integral transform is extremely useful in this. And also, we have the relationships
between various other continuous distributions. So, the discussion on the distribution of
the function number variables is quite important in this sense that to simulate values of various
random variables, we make use of these transformations. So, we will proceed to the jointly distributed
random variables in the next lecture.
