Let X and Y be jointly continuous nonnegative random variables. A particular value y of Y is observed and it turns out that fX|Y(x∣y)=2e−2x , for x≥0 .

Question

Let X and Y be jointly continuous nonnegative random variables. A particular value y of Y is observed and it turns out that fX|Y(x∣y)=2e−2x , for x≥0 .

Find the LMS estimate (conditional expectation) of X .
unanswered

Find the conditional mean squared error E[(X−XˆLMS)2∣Y=y] .
unanswered

Find the MAP estimate of X .
unanswered

Find the conditional mean squared error E[(X−XˆMAP)2∣Y=y] .
unanswered

Answer 1

a) 1/2

b) 1/4
c) 0
d) 1/2

Answer 2

Why did the random variable go to therapy? Because it was feeling marginal and needed some conditioning!

Answer 3

To find the LMS estimate (conditional expectation) of X, we need to find E[X|Y=y].

Using the conditional probability density function fX|Y(x|y), we can compute the conditional expectation as follows:

E[X|Y=y] = ∫x * fX|Y(x|y) dx

Since fX|Y(x|y) = 2e^(-2x) for x≥0, the conditional expectation becomes:

E[X|Y=y] = ∫x * 2e^(-2x) dx

To solve this integral, we can use integration by parts:

Let u = x and dv = 2e^(-2x) dx
Then, du = dx and v = -e^(-2x)

Using the integration by parts formula: ∫u dv = uv - ∫v du, we have:

∫x * 2e^(-2x) dx = -x * e^(-2x) - ∫(-e^(-2x)) dx

Simplifying the integral further:

= -x * e^(-2x) + ∫e^(-2x) dx
= -x * e^(-2x) - (1/2) * e^(-2x) + C

Now, evaluating the integral at the lower and upper limits:

∫x * 2e^(-2x) dx (from 0 to ∞) = [ (-x * e^(-2x) - (1/2) * e^(-2x)) ] (from 0 to ∞)

As x approaches infinity, e^(-2x) goes to 0, so the first term in the bracket also goes to 0.

Solving for the second term when x is 0:

= [ (-0 * e^0 - (1/2) * e^0) ] - [ (-0 * e^0 - (1/2) * e^0) ]
= 0 - (-(1/2))
= 1/2

Therefore, the LMS estimate (conditional expectation) of X given Y=y is 1/2.

Answer 4

To find the LMS estimate (conditional expectation) of X given that a particular value y of Y is observed, we need to calculate E[X|Y=y]. Let's begin by using the definition of conditional expectation:

E[X|Y=y] = ∫x fX|Y(x|y) dx

From the given information, we know that fX|Y(x|y) = 2e^(−2x) for x≥0. Now, let's substitute this into the formula:

E[X|Y=y] = ∫x (2e^(−2x)) dx

To evaluate this integral, we can use integration by parts. Let's denote the integral as I:

I = ∫x (2e^(−2x)) dx

Let u = x and dv = 2e^(−2x) dx, then du = dx and v = -e^(−2x)/2. Now, we can apply the integration by parts formula:

I = uv - ∫v du
I = - x(e^(−2x)/2) - ∫(-e^(−2x)/2) dx
I = - x(e^(−2x)/2) + (1/4) e^(−2x) + C

Finally, we can evaluate the definite integral from 0 to infinity to get the LMS estimate:

E[X|Y=y] = [ - x(e^(−2x)/2) + (1/4) e^(−2x) ] evaluated from 0 to infinity

By evaluating the limit as x approaches infinity, the second term goes to 0. Therefore, we are left with:

E[X|Y=y] = [ - x(e^(−2x)/2) ] evaluated from 0 to infinity
E[X|Y=y] = 0

Hence, the LMS estimate of X given Y=y is 0.

Answer 5

Can you show your calculations for a)?

I got a different result: 1/4

E[X|Y=y] = integral x*fX|Y(x|y)dx
= integral(infinity to 0) x*2e^(-2x) dx
=1/4

Am i wrong here?