Question

1. Suppose the random variable as a uniform distribution on [-k,k] a) Construct the pdf X. b) Calculate P(X > 2X > 1) in terms of 'k c) Calculate 'K if P(-2 < X < 2) =

Answer 1

Question (a)

The PDF for a continuous uniform distribution P(x) is constant over the possible values of x

P(x) = 1/ b-a for a$\leq$x$\leq$b

= 0 elsewhere

If the limits are a and b then the base of the rectangle will be of lenght b-a

We know that area of the rectangle = 1 (since the sum of all probabilities is 1)

So lenght of base * height = 1

Height = 1/b-a which is the PDF for values between a and b

The same is shown in below picture

o a - - - - - - b x

So here given the interval [-k,k]

So the PDF will be

P(X) = 1/k-(-k) for -k$\leq$X$\leq$k

= 1/2k for -k$\leq$X$\leq$k

So PDF is

P(X) = 1/2k for -k$\leq$X$\leq$k

= 0 elsehwere or for any other value of X

Question (b)

We need to find P(X>2 | X>1) in terms of k

Here we need to find the conditional probability, given that X>1 ix true we need to find the probability that X>2

SInce we already know that X>1, we will be no longer starting at -k, instead we will start at 1

So f(X) = 1/k-1 for 1$\leq$X$\leq$k

But we want the probability only for P(X>2) so the new base will be k-2

So P(X>2 | X-1) = base * height = (k-2) * (1/k-1) since height is PDF here

P(X>2 | X-1) = (k-2) / (k-1)

Question (c)

We need to find k if P(-2<X<2) = 1/5

Now our base will be 2-(-2) which will be 4 sicne X values ranges between -2 and 2

Height is still the same which is 1/2k

So P(-2<X<2) will be 4*(1/2k)

P(-2<X<2) = 2/k

Given that P(-2<X<2) = 1/5

so 2/k = 1/5

So k = 10