Exploring the CDF and MGF Relationship for Random Variables

[EngWiPy]

Hello,

Suppose that the Cumulative Distribution Function (CDF) of a random variable X is $$F_X(x)$$, which is by definition is:

$$F_X(x)=\text{Pr}\left[X\leq x\right]=\text{Pr}\left[\frac{1}{X}\geq \frac{1}{x}\right]=1-\text{Pr}\left[\frac{1}{X}\leq \frac{1}{x}\right]=1-F_{1/X}\left(1/x\right)$$

Considering this relation between the CDF of X and the CDF of its reciprocal, what is the relation between the Moment Generating Function (MGF) of X and its reciprocal?

Any help will be highly appreciated.

Thanks in advance

 

[Pere Callahan]

A good starting point would be to think of a relation between the CDF of X and the MGF of X, wouldn't it?

 

[EngWiPy]

A good starting point would be to think of a relation between the CDF of X and the MGF of X, wouldn't it?

Yes right, and I know what is the relation between them, but I want to see if another one has another idea. Anyway, the relation is:

$$M_X(s)=s\mathcal{L}\left\{F_X(x)\right\}$$

I have tried this, and it yields no where.

Regards

 

[bpet]

What would the CDF and MGF look like if X is uniform on [0,1] ?

 

[EngWiPy]

What would the CDF and MGF look like if X is uniform on [0,1] ?

The CDF of a uniformly distributed random variable X is:

$$F_X(x)=\begin{cases}0&x<0\\\frac{x-a}{b-a}&a\leq x<b\\1&x\ge b\end{cases}$$

Here, it may easier to derive the MGF from the PDF, not from the CDF. The PDF of X will be:

$$f_X(x)=\begin{cases}\frac{1}{b-a}&a\leq x\leq b\\0&\mbox{elsewhere}\end{cases}$$

Then the MGF of X is:

$$\mathcal{M}_X(s)=E_X\left[\text{e}^{sx}\right]=\int_a^b\text{e}^{sx}f_X(x)\,dx=\frac{\text{e}^{bs}-\text{e}^{as}}{s\left(b-a\right)}$$

But, what is the relation of this to the primary question?

Anyway, I have found the following relations between the MGF of X and the MGF of its reciprocal:

$$\mathcal{M}_X(s)=\int_s^{\infty}\int_0^{\infty}J_0\left(2\,\sqrt{u\,p}\right)\mathcal{M}_{1/X}(s)\,du\,dp\\$$
$$\mathcal{M}_{1/X}(s)=1-\sqrt{s}\int_0^{\pi/2}\frac{\sec^2 (\zeta)}{\sqrt{\tan (\zeta)}}J_1\left(2\,\sqrt{s\tan (\zeta)}\right)\mathcal{M}_X\left(\tan (\zeta)\right)\,d\zeta$$

where $$J_v(.)$$ is the vth order bessel function of the first kind. I do not know how they got there. Does anybody know how to derive these relations?

Thanks in advance

 

[mathman]

Hello,

Suppose that the Cumulative Distribution Function (CDF) of a random variable X is $$F_X(x)$$, which is by definition is:

$$F_X(x)=\text{Pr}\left[X\leq x\right]=\text{Pr}\left[\frac{1}{X}\geq \frac{1}{x}\right]=1-\text{Pr}\left[\frac{1}{X}\leq \frac{1}{x}\right]=1-F_{1/X}\left(1/x\right)$$

Considering this relation between the CDF of X and the CDF of its reciprocal, what is the relation between the Moment Generating Function (MGF) of X and its reciprocal?

Any help will be highly appreciated.

Thanks in advance

If X < 0 and x > 0, your statement about reciprocals doesn't hold.

 

[EngWiPy]

If X < 0 and x > 0, your statement about reciprocals doesn't hold.

Yes , I forgot to mention that $$0\leq X, x<\infty$$. Then, is there any problem?

Regards

 

[mathman]

Yes , I forgot to mention that $$0\leq X, x<\infty$$. Then, is there any problem?

Regards

Not in your original statement.