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By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +2\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +2\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +2\sigma dW^Q_t$$ wewe can deduce that ( $r = 0$) $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 $$$$V_0 = E^Q(X_T^2) = E^Q(x_0^2 e^{\sigma^2 T} e^{-\frac{1}{2}(2\sigma)^2 T+2\sigma W_T}) = x_0^2e^{\sigma^2 T} E^Q(e^{-\frac{1}{2}(2\sigma)^2 T+2\sigma W_T}) = x_0^2e^{\sigma^2 T} = e^{\sigma^2 T} $$

By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +2\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +2\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +2\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 $$

By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +2\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +2\sigma dW_t$$

we can deduce that ( $r = 0$) $$V_0 = E^Q(X_T^2) = E^Q(x_0^2 e^{\sigma^2 T} e^{-\frac{1}{2}(2\sigma)^2 T+2\sigma W_T}) = x_0^2e^{\sigma^2 T} E^Q(e^{-\frac{1}{2}(2\sigma)^2 T+2\sigma W_T}) = x_0^2e^{\sigma^2 T} = e^{\sigma^2 T} $$

The volatility of $X_t^2$ is $2\sigma"
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NN2
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By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +\sigma dW_t)$$$$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +2\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +\sigma dW_t$$$$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +2\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +\sigma dW^Q_t$$$$\frac{dX_t^2}{X_t^2} = r dt +2\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 $$$$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 $$

By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 $$

By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +2\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +2\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +2\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}(2\sigma)^2 (T-t)+2\sigma(W_T-W_t))}) = x_0^2 $$

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NN2
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By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}x_0^2 $$$$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 $$

By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}x_0^2 $$

By applying the Ito's lemma on $X_t^2$, you find easily the process $(X_t^2)$ satisfying $$d(X_t^2) = 2X_tdX_t +\frac{1}{2} 2 <dX_t,dX_t> = X_t^2 ( \sigma^2 dt +\sigma dW_t)$$

so $(X_t^2)$ is a geometric Brownian motion with drift $\mu = \sigma^2$ $$\frac{dX_t^2}{X_t^2} = \sigma^2 dt +\sigma dW_t$$

In risk neutral measure, the drift will be $r$, the process $X_t^2$ will become $$\frac{dX_t^2}{X_t^2} = r dt +\sigma dW^Q_t$$ we can deduce that $$V_t = e^{-(T-t)r}E^Q(X_T^2) = e^{-(T-t)r}E^Q(x_0^2e^{(T-t)r} e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 E^Q(e^{-\frac{1}{2}\sigma^2 (T-t)+(W_T-W_t))}) = x_0^2 $$

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