Question

A random variable Y is a function of random variable X, where y=x^2 and fx(x)=(x+1)/2 from -1 to 1 and =0 elsewhere. Determine fy(y). In this problem, there are two x values for every y value, which means x=T^-1(y)= +y^0.5 and -y^0.5. Be sure you account for both of these. Ans: fy(y)=0.5y^-0.5

Answer 1

Answer:-

Given That:-

A random variable Y is a function of random variable X, where y=x^2 and fx(x)=(x+1)/2 from -1 to 1 and =0 elsewhere. Determine fy(y). In this problem, there are two x values for every y value, which means x=T^-1(y)= +y^0.5 and -y^0.5. Be sure you account for both of these. Ans: fy(y)=0.5y^-0.5

Given that the pdf for X as

Let FY is the respective distribution function for Y and fY, is the corresponding pdf for Y.

Here, we can see that Y = X2 is not one-to-one function, For this, we need to find f, using generic method.

As we know,

[Y is the random variable, y is the realized value for Y]

i.e,

Therefore, using the relationship between pdf and its distribution function

We can have

i.e,

....................................(II)

Using the pdf of X in (II) we can have

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