Conditional Distribution: 7 Interesting Facts To Know

• Content
• Conditional distribution
• Discrete conditional distribution
• Example on discrete conditional distribution
• Continuous conditional distribution
• Example on Continuous conditional distribution
• Conditional distribution of bivariate normal distribution
• Joint Probability distribution of function of random variables
• Examples on Joint Probability distribution of function of random variables

Conditional distribution

   It is very interesting to discuss the conditional case of distribution when two random variables follows the distribution satisfying one given another, we first briefly see the conditional distribution in both the case of random variables, discrete and continuous then after studying some prerequisites we focus on the conditional expectations.

Discrete conditional distribution

     With the help of joint probability mass function in joint distribution we define conditional distribution for the discrete random variables X and Y using conditional probability for X given Y as the distribution with the probability mass function

provided the denominator probability is greater than zero, in similar we can write this as

in the joint probability if the X and Y are independent random variables then this will turn into

so the discrete conditional distribution or conditional distribution for the discrete random variables X given Y is the random variable with the above probability mass function in similar way for Y given X we can define.

Example on discrete conditional distribution

1. Find the probability mass function of random variable X given Y=1, if the joint probability mass function for the random variables X and Y has some values as

p(0,0)=0.4 , p(0,1)=0.2, p(1,0)= 0.1, p(1,1)=0.3

Now first of all for the value Y=1 we have

so using the definition of probability mass function

we have

and

• obtain the conditional distribution of X given X+Y=n, where X and Y are Poisson distributions with the parameters λ₁ and λ₂ and X and Y are independent random variables

Since the random variables X and Y are independent, so the conditional distribution will have probability mass function as

since the sum of Poisson random variable is again poisson so

thus the conditional distribution with above probability mass function will be conditional distribution for such Poisson distributions. The above case can be generalize for more than two random variables.

Continuous conditional distribution

   The Continuous conditional distribution of the random variable X given y already defined is the continuous distribution with the probability density function

denominator density is greater than zero, which  for the continuous density function is

thus the probability for such conditional density function is

In similar way as in discrete if X and Y are independent  in continuous then also

and hence

so we can write it as

Example on Continuous conditional distribution

1. Calculate conditional density function of random variable X given Y if the joint probability density function with the open interval (0,1) is given by

If for the random variable X given Y within (0,1) then by using the above density function we have

• Calculate the conditional probability

if the joint probability density function is given by

To find the conditional probability first we require the conditional density function so by the definition it would be

now using this density function in the probability the conditional probability is

Conditional distribution of bivariate normal distribution

  We know that the Bivariate normal distribution of the normal random variables X and Y with the respective means and variances as the parameters has the joint probability density function

so to find the conditional distribution for such a bivariate normal distribution for X given Y is defined by following the conditional density function of the continuous random variable and the above joint density function we have

By observing this we can say that this is normally distributed with the mean

and variance

in the similar way the conditional density function for Y given X already defined will be just interchanging the positions of the parameters of X with Y,

The marginal density function for X we can obtain from the above conditional density function by using the value of the constant

let us substitute in the integral

the density function will be now

since the total value of

by the definition of the probability so the density function will be now

which is nothing but the density function of random variable X with usual mean and variance as the parameters.

Joint Probability distribution of function of random variables

  So far we know the joint probability distribution of two random variables, now if we have functions of such random variables then what would be the joint probability distribution of those functions, how to calculate the density and distribution function because we have real life situations where we have functions of the random variables,

If Y₁ and Y₂ are the functions of the random variables X₁ and X₂ respectively which are jointly continuous then the joint continuous density function of these two functions will be

where Jacobian

and Y₁ =g₁ (X₁, X₂) and Y₂ =g₂ (X₁, X₂) for some functions g₁ and g₂ . Here g₁ and g₂ satisfies the conditions of the Jacobian as continuous and have continuous partial derivatives.

Now the probability for such functions of random variables will be

Examples on Joint Probability distribution of function of random variables

1. Find the joint density function of the random variables Y₁ =X₁ +X₂ and Y₂=X₁ -X₂ , where X₁ and X₂ are the jointly continuous with joint probability density function. also discuss for the different nature of distribution .

Here we first we will check Jacobian

since g₁(x₁, x₂)= x₁ + x₂  and g₂(x₁, x₂)= x₁ – x₂ so

simplifying Y₁ =X₁ +X₂ and Y₂=X₁ -X₂ , for the value of X₁ =1/2( Y₁ +Y₂ ) and X₂ = Y₁ -Y₂ ,

if these random variables are independent uniform random variables

or if these random variables are independent exponential random variables with usual parameters

or if these random variables are independent normal random variables then

• If X and Y are the independent standard normal variables as given

calculate the joint distribution for the respective polar coordinates.

We will convert by usual conversion X and Y into r and θ as

so the partial derivatives of these function will be

so the Jacobian using this functions is

if both the random variables X and Y are greater than zero then conditional joint density function is

now the conversion of cartesian coordinate to the polar coordinate using

so the probability density function for the positive values will be

for the different combinations of X and Y the density functions in similar ways are

now from the average of the above densities we can state the density function as

and the marginal density function from this joint density of polar coordinates over the interval (0, 2π)

• Find the joint density function for the function of random variables

U=X+Y and V=X/(X+Y)

where X and Y are the gamma distribution with parameters (α + λ) and (β +λ) respectively.

Using the definition of gamma distribution and joint distribution function the density function for the random variable X and Y will be

consider the given functions as

g₁ (x,y) =x+y , g₂ (x,y) =x/(x+y),

so the differentiation of these function is

now the Jacobian is

after simplifying the given equations the variables x=uv and y=u(1-v) the probability density function is

we can use the relation

• Calculate the joint probability density function for

Y₁ =X₁ +X₂+ X₃ , Y₂ =X₁– X₂ , Y₃ =X₁ – X₃

where the random variables X1 , X2, X3 are the standard normal random variables.

Now let us calculate the Jacobian by using partial derivatives of

Y₁ =X₁ +X₂+ X₃ , Y₂ =X₁– X₂ , Y₃ =X₁ – X₃

as

simplifying for variables X₁ , X₂ and X₃

X₁ = (Y₁ + Y₂ + Y₃)/3 , X₂ = (Y₁ – 2Y₂ + Y₃)/3 , X₃ = (Y₁ + Y₂ -2 Y₃)/3

we can generalize the joint density function as

so we have

for the normal variable the  joint probability density function is

hence

where the index is

compute the joint density function of Y₁ ……Y_n and marginal density function for Y_n where

and X_i are independent identically distributed exponential random variables with parameter λ.

for the random variables of the form

Y₁ =X₁ , Y₂ =X₁ + X₂ , ……, Y_n =X₁ + ……+ X_n

the Jacobian will be of the form

and hence its value is one, and the joint density function for the exponential random variable

and the values of the variable X_i ‘s will be

so the joint density function is

Now to find the marginal density function of Y_n we will integrate one by one  as

and

like wise

if we continue this process we will get

which is the marginal density function.

Conclusion:

The conditional distribution for the discrete and continuous random variable with different examples considering some of the types of these random variables discussed, where the independent random variable plays important role. In addition the  joint distribution for the function of joint continuous random variables also explained with suitable examples, if you require further reading go through below links.

For more post on Mathematics, please refer to our Mathematics Page

Wikipediahttps://en.wikipedia.org/wiki/joint_probability_distribution/” target=”_blank” rel=”noreferrer noopener” class=”rank-math-link”>Wikipedia.org

A first course in probability by Sheldon Ross

Schaum’s Outlines of Probability and Statistics

An introduction to probability and statistics by ROHATGI and SALEH