# Generating PDF from random samples

#### michellerenee

##### New Member
Hello,

I am taking an online course and have no instructor to help me....HELP!

Suppose the random variable X has the pdf f(x)=e^(-x). Let the random variable Y= X(1) + X(2) where X(1) and X(2) are random samples from the above pdf. *Note: the 1 and 2 that follow the X's are in subscript form.

1) Find the pdf of Y.
2) What is the expected value of Y?
3) What is the probability that Y will exceed 1?

I do not even have a clue where to start or I would put out what I think.
Thank you!

#### Martingale

##### TS Contributor
Hello,

I am taking an online course and have no instructor to help me....HELP!

Suppose the random variable X has the pdf f(x)=e^(-x). Let the random variable Y= X(1) + X(2) where X(1) and X(2) are random samples from the above pdf. *Note: the 1 and 2 that follow the X's are in subscript form.

1) Find the pdf of Y.
2) What is the expected value of Y?
3) What is the probability that Y will exceed 1?

I do not even have a clue where to start or I would put out what I think.
Thank you!
I'll use X_1 for X(1)...

1) if X_1 and X_2 are a random sample each with pdf f(x) then you can find the distribution of Y=X_1+X_2 by using convolutions.

2) E(Y)=E(X_1)+E(X_2)=2*E(X_1) since X_1 and X_2 have the same distribution. (or you can use the pdf you found in (1) to get the answer)

3) P(Y>=1)=... use the distribution you found in (1)

#### Mean Joe

##### TS Contributor
1) Find the pdf of Y.
Usually, when you want to find the pdf, your first step will be to find the cdf. Which you can do using this:
F(y) = P[Y <= y]

Since Y = X[1] + X[2], then (let's reason together) for Y <= y we need X[1] <= y and X[2] <= y - x1. Thus
F(y) = P[ X[1] <= y and X[2] <= y - x1 ]

Now you get to do calculus, since the random variables X[1] and X[2] are continuous:
F(y) = P[ ... ] = Integral(0 to y, Integral(0 to y-x1, e^(-x2) dx2) * e^(-x1) dx1)

The notation I'm using above puts two parameters into the integral function: Integral( limits, integrand). You have two random variables, so you have a double integral.

#### Mean Joe

##### TS Contributor
I'll use X_1 for X(1)...

1) if X_1 and X_2 are a random sample each with pdf f(x) then you can find the distribution of Y=X_1+X_2 by using convolutions.
Question about convolution: I tried using this formula, from wikipedia:

I get f[Y] (y) = f[X1 + X2] (y) = Integral(0 to infinity, f[X1](z)*f[X2](y-z) dz)
= Integral(0 to infinity, e^-(z) * e^-(y-z) dz)
= Integral(0 to infinity, e^(-y) dz)
does not exist??

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#### Martingale

##### TS Contributor
Question about convolution: I tried using this formula, from wikipedia:

I get f[Y] (y) = f[X1 + X2] (y) = Integral(0 to infinity, f[X1](z)*f[X2](y-z) dz)
= Integral(0 to infinity, e^-(z) * e^-(y-z) dz)
= Integral(0 to infinity, e^(-y) dz)
does not exist??
if y<z then

y-z<0

thus f[x2](y-z)=0